diff options
Diffstat (limited to '1092')
194 files changed, 10408 insertions, 0 deletions
diff --git a/1092/CH1/EX1.1/Example1_1.sce b/1092/CH1/EX1.1/Example1_1.sce new file mode 100755 index 000000000..a07815d67 --- /dev/null +++ b/1092/CH1/EX1.1/Example1_1.sce @@ -0,0 +1,22 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+t = 50e-3; // t = time in milli second
+phi = 8 * 10 ^ 6; // phi = uniform magnetic field in maxwells
+
+// Calculations
+E_av = (phi / t) * 10 ^ -8; // E_av = average voltage generated in the conductor
+// in volt
+
+// Display the result
+disp("Example 1-1 Solution : ");
+disp("Average voltage generated in the conductor is : ");
+printf(" E_av = %.2f V", E_av);
diff --git a/1092/CH1/EX1.10/Example1_10.sce b/1092/CH1/EX1.10/Example1_10.sce new file mode 100755 index 000000000..5502597d0 --- /dev/null +++ b/1092/CH1/EX1.10/Example1_10.sce @@ -0,0 +1,28 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+no_of_coils = 40;
+N = 20; // no of turns in each coil
+omega = 200; // angular velocity of armature in rad/s
+phi = 5 * 10 ^ -3; // flux per pole
+a = 4; // No. of parallel paths
+P = 4; // No. of poles
+
+// Calculations
+Z = no_of_coils * 2 * N; // No. of conductors
+
+E_g = ( phi * Z * omega * P ) / ( 2 * %pi * a ); // Voltage generated by the
+// armature between brushes
+
+// Display the results
+disp("Example 1-10 Solution : ");
+printf("\n Z = % d conductors ", Z);
+printf("\n Eg = % .2f V between the brushes ", E_g);
diff --git a/1092/CH1/EX1.11/Example1_11.sce b/1092/CH1/EX1.11/Example1_11.sce new file mode 100755 index 000000000..2853b6621 --- /dev/null +++ b/1092/CH1/EX1.11/Example1_11.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+l = 0.5; // length of the conductor
+A = 0.1 * 0.2; // area of the pole face
+phi = 0.5 * 10 ^ -3; // magnetic flux in weber
+I = 10; //Current in the conductor
+
+// Calculations
+B = ( phi ) / ( A ); // Flux density
+
+F = B * I * l; // Magnitude of force
+
+// Display the result
+disp("Example 1-11 Solution : ");
+
+printf("\n a : F = % .3f N", F );
+
+printf("\n b : The force on the conductor is % .3f N in an upward direction as shown in fig 1-13c ", F );
diff --git a/1092/CH1/EX1.12/Example1_12.sce b/1092/CH1/EX1.12/Example1_12.sce new file mode 100755 index 000000000..87398b045 --- /dev/null +++ b/1092/CH1/EX1.12/Example1_12.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+l = 0.5; // length of the conductor
+A = 0.1 * 0.2; // area of the pole face
+phi = 0.5 * 10 ^ -3; // magnetic flux in weber
+I = 10; //Current in the conductor
+theta = 75; // angle between the conductor and the flux density B
+
+// Calculations
+B = ( phi ) / ( A ); // Flux density
+
+F = B * I * l * sind(theta); // Magnitude of force
+
+// Display the result
+disp("Example 1-12 Solution : ");
+
+printf("\n F =% f N in a vertically upward direction ", F );
+
diff --git a/1092/CH1/EX1.13/Example1_13.sce b/1092/CH1/EX1.13/Example1_13.sce new file mode 100755 index 000000000..e5775f743 --- /dev/null +++ b/1092/CH1/EX1.13/Example1_13.sce @@ -0,0 +1,21 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-13
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_a = 0.25; // Armature resistance
+V_a = 125; // dc bus voltage
+I_a = 60; // Armature current
+
+// Calculations
+E_c = V_a - I_a * R_a; // Counter EMF generated in the armature conductors of motor
+
+// Display the result
+disp("Example 1-13 Solution : ");
+printf("\n Ec = % d V ", E_c );
diff --git a/1092/CH1/EX1.14/Example1_14.sce b/1092/CH1/EX1.14/Example1_14.sce new file mode 100755 index 000000000..640b6fffc --- /dev/null +++ b/1092/CH1/EX1.14/Example1_14.sce @@ -0,0 +1,35 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-13
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_a = 110; // voltage across armature
+I_a = 60; // Armature current
+R_a = 0.25; // Armature resistance
+P = 6; // No. of poles
+a = 12; // No. of paths
+Z = 720; // No. of armature conductors
+S = 1800; // Speed in rpm
+
+// Calculations
+E_g = V_a + I_a * R_a; // Generated EMF in the armature
+
+phi_lines = ( E_g * ( 60 * a ) ) / ( ( Z * S * P ) * 10 ^ -8 );
+// Flux per pole in lines
+
+phi_Wb = phi_lines * 10 ^ -8; // Flux per pole in webers
+
+// Display the results
+disp("Example 1-14 Solution : ");
+
+printf("\n a : Eg = %d V ", E_g );
+
+printf("\n b : phi = %f lines/pole ", phi_lines );
+
+printf("\n c : phi = %f Wb ", phi_Wb );
diff --git a/1092/CH1/EX1.2/Example1_2.sce b/1092/CH1/EX1.2/Example1_2.sce new file mode 100755 index 000000000..6d3ed6f45 --- /dev/null +++ b/1092/CH1/EX1.2/Example1_2.sce @@ -0,0 +1,31 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+l = 18; // l = length of the conductor in inches
+B = 50000; // B = uniform magnetic field in lines/sq-inches
+d = 720; // d = distance travelled by conductor in inches
+t = 1; // t =time taken for the conductor to move in second
+
+// Calculations
+v = d/t; // v = velocity in inches/second with which the conductor moves
+
+// part a
+e = B * l * v * 10 ^ -8; // e = instantaneous induced EMF in volt
+// part b
+A = d * l; // Area swept by the conductor while moving
+phi = B * A; // phi = uniform magnetic field
+E = ( phi / t ) * 10 ^ -8; // E = average induced EMF
+
+// Display the result
+disp("Example 1-2 Solution : ");
+
+printf(" \n a : e = %.2f V ", e);
+printf(" \n b : E = %.2f V ", E);
diff --git a/1092/CH1/EX1.3/Example1_3.sce b/1092/CH1/EX1.3/Example1_3.sce new file mode 100755 index 000000000..da791c32c --- /dev/null +++ b/1092/CH1/EX1.3/Example1_3.sce @@ -0,0 +1,28 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+l = 18; // l = length of the conductor in inches
+B = 50000; // B = uniform magnetic field in lines/sq-inches
+d = 720; // d = distance travelled by conductor in inches
+t = 1; // t =time taken for the conductor to move in second
+theta = 75 // theta = angle between the motion of the conductor and field
+// in radians
+
+// Calculations
+v = d/t; // v = velocity in inches/second with which the conductor moves
+
+E = B * l * v * 10 ^ -8 * sind(theta); // E = Average induced EMF in volt
+
+// Display the result
+disp("Example 1-3 Solution : ");
+
+disp(" Average induced EMF in volt is :")
+printf(" E = %.2f V ", E);
diff --git a/1092/CH1/EX1.4/Example1_4.sce b/1092/CH1/EX1.4/Example1_4.sce new file mode 100755 index 000000000..ec06a7ce8 --- /dev/null +++ b/1092/CH1/EX1.4/Example1_4.sce @@ -0,0 +1,29 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+v = 1.5; // v = velocity in m/s with which the conductor is moving
+l = 0.4; // l = length of the conductor
+B = 1; // B = uniform field intensity in tesla
+theta_a = 90; // theta_a = angle between the motion of the conductor and field
+theta_b = 35; // theta_b = angle between the motion of the conductor and field
+theta_c = 120; // theta_c = angle between the motion of the conductor and field
+
+// Calculations
+E_a = B * l * v * sind(theta_a); // Voltage induced in the conductor for theta_a
+E_b = B * l * v * sind(theta_b); // Voltage induced in the conductor for theta_b
+E_c = B * l * v * sind(theta_c); // Voltage induced in the conductor for theta_c
+
+// Display the result
+disp("Example 1-1 Solution : ");
+
+printf("\n a: E = %.2f V ", E_a);
+printf("\n b: E = %.3f V ", E_b);
+printf("\n c: E = %.2f V ", E_c);
diff --git a/1092/CH1/EX1.5/Example1_5.sce b/1092/CH1/EX1.5/Example1_5.sce new file mode 100755 index 000000000..b167d0475 --- /dev/null +++ b/1092/CH1/EX1.5/Example1_5.sce @@ -0,0 +1,53 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+no_of_conductors = 40;
+A = 2; // A = Parallel paths
+path = A;
+flux_per_pole = 6.48 * 10 ^ 8; // flux lines
+S = 30; // S = Speed of the prime mover in rpm
+R_per_path = 0.01; // Resistance per path
+I = 10; // Current carried by each condutcor
+P = 2; // No. of poles
+
+// Calculations
+total_flux = P * flux_per_pole; // Total flux linked in one revolution
+t = ( 1 / 30 ) * ( 60 ); // time for one revolution
+
+e_av_per_conductor = ( total_flux / t ) * 10^-8; // Average voltage generated
+// per conductor
+E_path = ( e_av_per_conductor ) * ( no_of_conductors / path ); // Average
+//voltage generated per path
+
+E_g = E_path; // Generated armature voltage
+
+I_a =( I / path ) * ( 2 * path ); // Armature current delivered to an external
+// load
+
+R_a = ( R_per_path) / path * 20; // Armature resistance
+
+V_t = E_g - I_a * R_a; // Terminal voltage of generator
+
+P = V_t * I_a; // Genrator power rating
+
+// Display the results
+disp("Example 1-5 Solution");
+
+printf(" \n a : E/path = %.2f V/path ", E_path );
+printf(" \n b : Eg = %.2f V ", E_g );
+printf(" \n c : Ia = %.2f A ", I_a );
+printf(" \n d : Ra = %.2f ohm ", R_a );
+printf(" \n e : Vt = %.2f V ", V_t );
+printf(" \n f : P = %.2f W ", P );
+
+
+
+
diff --git a/1092/CH1/EX1.6/Example1_6.sce b/1092/CH1/EX1.6/Example1_6.sce new file mode 100755 index 000000000..183d91088 --- /dev/null +++ b/1092/CH1/EX1.6/Example1_6.sce @@ -0,0 +1,45 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+no_of_conductors = 40;
+I = 10; // Current carried by each condutcor
+R_per_path = 0.01; // Resistance per path
+flux_per_pole = 6.48 * 10 ^ 8; // flux lines
+P = 2; // No. of poles
+path = 4; // No. of parallel paths
+total_flux = P * flux_per_pole; // Total flux linked in one revolution
+t = 2; // time for one revolution
+e_av_per_conductor = 6.48; // Average voltage generated per conductor
+
+// Calculations
+E_path = ( e_av_per_conductor ) * ( no_of_conductors / path ); // Average
+//voltage generated per path
+
+E_g = E_path; // Generated armature voltage
+
+I_a =( I / path ) * ( 4 * path ); // Armature current delivered to an external
+// load
+
+R_a = ( ( R_per_path) / path ) * 10; // Armature resistance
+
+V_t = E_g - I_a * R_a; // Terminal voltage of generator
+
+P = V_t * I_a; // Genrator power rating
+
+// Display the results
+disp("Example 1-6 Solution");
+
+printf(" \n a : E/path = %.2f V/path ", E_path );
+printf(" \n b : Eg = %.2f V ", E_g );
+printf(" \n c : Ia = %.2f A ", I_a );
+printf(" \n d : Ra = %.3f ohm ", R_a );
+printf(" \n e : Vt = %.2f V ", V_t );
+printf(" \n f : P = %.2f W ", P );
diff --git a/1092/CH1/EX1.7/Example1_7.sce b/1092/CH1/EX1.7/Example1_7.sce new file mode 100755 index 000000000..9686edab9 --- /dev/null +++ b/1092/CH1/EX1.7/Example1_7.sce @@ -0,0 +1,25 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+N = 1; // no. of turns
+phi = 6.48 * 10 ^ 8; // Magnetic flux in lines
+s = 30 / 60; // No. of revolution of the coil per second( refer section 1-14)
+
+// Calculations
+E_av_per_coil = 4 * phi * N * s * 10 ^ -8; // average voltage per coil
+// for above equation refer section 1-14
+
+E_av_per_coil_side = E_av_per_coil * ( 1 / 2); // average voltage per conductor
+
+// Display the results
+disp("Example 1-7 Solution : ")
+printf(" \n Eav/coil = % .2f V/coil ", E_av_per_coil);
+printf(" \n Eav/coil side = % .2f V/conductor ", E_av_per_coil_side);
diff --git a/1092/CH1/EX1.8/Example1_8.sce b/1092/CH1/EX1.8/Example1_8.sce new file mode 100755 index 000000000..5b12a448e --- /dev/null +++ b/1092/CH1/EX1.8/Example1_8.sce @@ -0,0 +1,25 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+phi_lines = 6.48 * 10 ^ 8; // magnetic flux in lines
+N = 1; // no. of turns
+
+// Calculations
+phi = phi_lines * 10 ^ -8; // Magnetic flux in weber
+
+omega = ( 30 ) * ( 2 * %pi ) * ( 1 / 60 ); // angular velocity in rad/s
+
+E_av_per_coil = 0.63662 * omega * phi * N; // average voltage per coil
+// for the above formula refer section 1-14 eqn (1-4b)
+
+// Display the result
+disp("Example 1-8 Solution : ");
+printf("\n Eav/coil = % 0.2f V/coil ", E_av_per_coil);
diff --git a/1092/CH1/EX1.9/Example1_9.sce b/1092/CH1/EX1.9/Example1_9.sce new file mode 100755 index 000000000..9f59f2427 --- /dev/null +++ b/1092/CH1/EX1.9/Example1_9.sce @@ -0,0 +1,24 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 1: Electromechanical Fundamentals
+// Example 1-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 2; // No. of poles
+Z = 40; // no of conductors
+a = 2; // a = Parallel paths
+phi = 6.48 * 10 ^ 8; // magnetic flux
+S = 30; // Speed of the prime mover
+
+// Calculations
+E_g = ( ( phi * Z * S * P ) / ( 60 * a) ) * 10 ^ -8; // average voltage between
+// the brushes
+
+// Display the result
+disp("Example 1-9 Solution : ");
+printf("\n Eg = %.2f V between the brushes ", E_g);
diff --git a/1092/CH10/EX10.1/Example10_1.sce b/1092/CH10/EX10.1/Example10_1.sce new file mode 100755 index 000000000..9230fe70d --- /dev/null +++ b/1092/CH10/EX10.1/Example10_1.sce @@ -0,0 +1,61 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 10: SINGLE-PHASE MOTORS
+// Example 10-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+hp = 0.25 ; // Power rating of the single-phase motor in hp
+V = 110 ; // Voltage rating of the single-phase motor in V
+I_sw = 4 ; // Starting winding current
+phi_I_sw = 15 ; // Phase angle in degrees by which I_sw lags behind V
+I_rw = 6 ; // Running winding current
+phi_I_rw = 40 ; // Phase angle in degrees by which I_rw lags behind V
+
+// Calculations
+// case a
+I_s = I_sw * exp( %i * -phi_I_sw*(%pi/180) ); // starting current in A
+// (%pi/180) for degrees to radians conversion of phase angle
+I_s_m = abs(I_s);//I_s_m = magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+I_r = I_rw * exp( %i * -phi_I_rw*(%pi/180) ); // running current in A
+I_r_m = abs(I_r);//I_r_m = magnitude of I_r in A
+I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees
+
+I_t = I_s + I_r ; // Total starting current in A
+I_t_m = abs(I_t);//I_t_m = magnitude of I_t in A
+I_t_a = atan(imag(I_t) /real(I_t))*180/%pi;//I_t_a=phase angle of I_t in degrees
+Power_factor = cosd(I_t_a); // Power factor
+
+// case b
+Is_cos_theta = real(I_s); // Component of the starting winding current in phase
+// with the supply voltage in A
+
+// case c
+Ir_sin_theta = imag(I_r); // Component of the running winding current that lags
+// the supply voltage by 90 degrees
+
+// case d
+phase = ( phi_I_rw - phi_I_sw ); // Phase angle between the starting and running
+// currents in degrees
+
+// Display the results
+disp("Example 10-1 Solution : ");
+printf(" \n a: I_s = %d <-%d A ", I_sw , phi_I_sw );
+printf(" \n I_s in A = " );disp(I_s);
+printf(" \n I_r = %d <-%d A ", I_rw , phi_I_rw );
+printf(" \n I_r in A = " );disp(I_r);
+printf(" \n I_t in A = " );disp(I_t);
+printf(" \n I_t = %.2f <%d A ", I_t_m , I_t_a );
+printf(" \n\n Power factor = cos(%d) = %.3f lagging \n", I_t_a ,Power_factor);
+
+printf(" \n b: Is*cosθ = %.2f A (from a)\n ", Is_cos_theta );
+
+printf(" \n c: (from a),\n Ir*sinθ in A = " );disp(%i*Ir_sin_theta);
+
+printf(" \n d: (θ_r - θ_s) = %d degrees ", phase);
diff --git a/1092/CH10/EX10.2/Example10_2.sce b/1092/CH10/EX10.2/Example10_2.sce new file mode 100755 index 000000000..739cb0ec9 --- /dev/null +++ b/1092/CH10/EX10.2/Example10_2.sce @@ -0,0 +1,45 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 10: SINGLE-PHASE MOTORS
+// Example 10-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data as per Ex.10-1
+hp = 0.25 ; // Power rating of the single-phase motor in hp
+V = 110 ; // Voltage rating of the single-phase motor in V
+I_s = 4 ; // Starting winding current
+phi_I_s = 15 ; // Phase angle in degrees by which I_s lags behind V
+I_r = 6 ; // Running winding current
+phi_I_r = 40 ; // Phase angle in degrees by which I_r lags behind V
+
+// Calculations
+// case a
+P_s = V * I_s * cosd(phi_I_s); // Power dissipated in the starting winding in W
+
+// case b
+P_r = V * I_r * cosd(phi_I_r); // Power dissipated in the running winding in W
+
+// case c
+P_t = P_s + P_r ; // Total instantaneous power dissipated during starting in W
+
+// case d
+P_r_d = P_r ; // Total steady-state power dissipated during running in W
+
+// case e
+eta = ( hp * 746 ) / P_r * 100 ; // Motor efficiency in percent
+
+// Display the results
+disp("Example 10-2 Solution : ");
+printf(" \n a: Power dissipated in the starting winding\n P_s = %d W \n", P_s );
+
+printf(" \n b: Power dissipated in the running winding\n P_r = %.1f W \n", P_r );
+
+printf(" \n c: Total instantaneous power dissipated during starting\n P_t = %.1f W \n", P_t );
+
+printf(" \n d: Total steady-state power dissipated during running\n P_r = %.1f W \n", P_r_d );
+
+printf(" \n e: Motor efficiency \n η = %.f percent \n", eta );
diff --git a/1092/CH10/EX10.3/Example10_3.sce b/1092/CH10/EX10.3/Example10_3.sce new file mode 100755 index 000000000..023a3774f --- /dev/null +++ b/1092/CH10/EX10.3/Example10_3.sce @@ -0,0 +1,67 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 10: SINGLE-PHASE MOTORS
+// Example 10-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+hp = 0.25 ; // Power rating of the single-phase motor in hp
+V = 110 ; // Voltage rating of the single-phase motor in V
+I_sw = 4 ; // Starting winding current
+phi_I_sw = 15 ; // Phase angle in degrees by which I_sw lags behind V
+I_rw = 6 ; // Running winding current
+phi_I_rw = 40 ; // Phase angle in degrees by which I_rw lags behind V
+// when the capacitor is added to the auxiliary starting winding of the motor
+// of Ex.10-1 , I_s leads V by 42 degrees so,
+phi_I_sw_new = 42 ; // I_s leads V by phi_I_sw_new degrees
+
+// Calculations
+// case a
+I_s = I_sw * exp( %i * phi_I_sw_new*(%pi/180) ); // starting current in A
+// (%pi/180) for degrees to radians conversion of phase angle
+I_s_m = abs(I_s);//I_s_m = magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+I_r = I_rw * exp( %i * -phi_I_rw*(%pi/180) ); // running current in A
+I_r_m = abs(I_r);//I_r_m = magnitude of I_r in A
+I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees
+
+I_t = I_s + I_r ; // Total starting current in A
+I_t_m = abs(I_t);//I_t_m = magnitude of I_t in A
+I_t_a = atan(imag(I_t) /real(I_t))*180/%pi;//I_t_a=phase angle of I_t in degrees
+Power_factor = cosd(I_t_a); // Power factor
+
+// case b
+theta = ( phi_I_rw - (-phi_I_sw_new) );
+sin_theta = sind(theta);// Sine of the angle between the
+// starting and running currents
+phase = 25 ; // Phase angle between the starting and running
+// currents in degrees (from Ex.10-1)
+
+// case c
+// Ratio of starting torques (capacitor to resistance start)
+ratio_T = sind(theta) / sind(phase);
+
+// Display the results
+disp("Example 10-3 Solution : ");
+printf(" \n a: I_s = %d <%d A ", I_sw , phi_I_sw_new );
+printf(" \n I_s in A = " );disp(I_s);
+printf(" \n I_r = %d <-%d A ", I_rw , phi_I_rw );
+printf(" \n I_r in A = " );disp(I_r);
+printf(" \n I_t in A = " );disp(I_t);
+printf(" \n I_t = %.2f <%.1f A ", I_t_m , I_t_a );
+printf(" \n\n Power factor = cos(%.1f) = %.3f lagging \n", I_t_a ,Power_factor);
+
+printf(" \n b: sin(%d - (-%d)) = sin(%d) = %.4f\n",phi_I_rw,phi_I_sw_new,theta,sin_theta);
+
+printf(" \n c: The steady state starting current has been reduced from");
+printf(" \n 9.77 <-30 A to %.2f <%.1f A ,",I_t_m ,I_t_a );
+printf(" \n and the power factor has risen from 0.866 lagging to %.3f.",Power_factor);
+printf(" \n The motor develops maximum starting torque(T = K*I_b*ϕ*cosθ) with");
+printf(" \n minimum starting current.The ratio of starting torques ");
+printf(" \n (capacitor to resistance start) is : \n");
+printf(" \n T_cs/T_rs = sin(%d)/sin(%d) = %.3f",theta,phase,ratio_T)
diff --git a/1092/CH10/EX10.4/Example10_4.sce b/1092/CH10/EX10.4/Example10_4.sce new file mode 100755 index 000000000..7528cdaee --- /dev/null +++ b/1092/CH10/EX10.4/Example10_4.sce @@ -0,0 +1,53 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 10: SINGLE-PHASE MOTORS
+// Example 10-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data (from Table 10-2)
+T_r = 1 ; // Rated torque in lb-ft
+T_s = 4.5 ; // Starting torque in lb-ft (rfom Locked-Rotor Data)
+T_br = 2.5 ; // Breakdown torque in lb-ft (Breakdown-Torque Data)
+
+// Rated Load Data
+P = 400 ; // Rated input power in W
+V = 115 ; // Rated input voltage in volt
+I_t = 5.35 ; // Rated input current in A
+Speed = 1750 ; // Rated speed in rpm
+
+// Calculations
+// case a
+ratio_s_r_T = T_s / T_r ; // Ratio of starting to rated torque
+
+// case b
+ratio_s_br_T = T_br / T_r ; // Ratio of breakdown to rated torque
+
+// case c
+P_o_hp = 1 / 3 ; // Power output in hp
+P_o = P_o_hp * 746 ; // Power output in W
+eta = P_o / P * 100 ; // Rated load efficiency
+
+// case d
+S = V * I_t ; // VA rating of the motor
+cos_theta = P / S ; // Rated load - power factor
+
+// case e
+T = 1 ; // Rated load torque in lb-ft
+hp = (T*Speed)/5252 ; // Rated load horsepower
+
+// Display the results
+disp("Example 10-4 Solution : ");
+
+printf(" \n a: T_s/T_r = %.1f \n ",ratio_s_r_T );
+
+printf(" \n b: T_br/T_r = %.1f \n ",ratio_s_br_T );
+
+printf(" \n c: Rated load efficiency \n η = %.1f percent \n ",eta );
+
+printf(" \n d: Rated load power factor\n cosθ = %.4f \n ",cos_theta );
+
+printf(" \n e: Rated load horsepower\n hp = %.4f hp ",hp);
diff --git a/1092/CH11/EX11.1/Example11_1.sce b/1092/CH11/EX11.1/Example11_1.sce new file mode 100755 index 000000000..1dcceb2f6 --- /dev/null +++ b/1092/CH11/EX11.1/Example11_1.sce @@ -0,0 +1,69 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 11: SPECIALIZED DYNAMOS
+// Example 11-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Torque - speed relations shown in Fig.11-3b for a dc servomotor.
+
+// Calculations
+// case a
+// Extrapolating to load line point x,
+S = 800 ; // Motor speed at point x
+V = 60 ; // Armature voltage in volt at point x
+
+// case b
+// At standstill, 60 V yields 4.5 lb-ft of starting torque
+T = 4.5 ;
+
+// case c
+P_c = (T*S)/5252 ; // Power delivered to the load in hp (from case a conditions)
+P_c_watt = P_c * 746 ; // P_c in W
+// case d
+// At point o:
+T_d = 1.1 ; // Starting torque in lb-ft (subscript d indicates case d) and
+S_d = 410 ; // Motor speed at point at point o
+
+// case e
+// At point w:
+T_e = 2.4 ; // Starting torque in lb-ft (subscript e indicates case e) and
+S_e = 900 ; // Motor speed at point at point w
+
+// case f
+P_d = (T_d*S_d)/5252 ; // Power delivered to the load in hp (from case d conditions)
+P_d_watt = P_d * 746 ; // P_d in W
+
+// case g
+P_f = (T_e*S_e)/5252 ; // Power delivered to the load in hp (from case f conditions)
+P_f_watt = P_f * 746 ; // P_f in W
+
+// case h
+// Upper limit of power ranges A and B are:
+A = 65 ; // Upper limit of power range A in W
+B = 305 ; // Upper limit of power range B in W
+
+// Display the results
+disp("Example 11-1 Solution : ");
+
+printf(" \n a: Extrapolating to load line point x,\n S = %d rpm ",S);
+printf(" \n Load line voltage is %d V \n",V);
+
+printf(" \n b: At standstill, %d V yields T = %.1f lb-ft of starting torque\n",V,T);
+
+printf(" \n c: Power delivered to the load in hp (from case a conditions)");
+printf(" \n P = %.4f hp = %d W \n",P_c,P_c_watt);
+
+printf(" \n d: At point o:\n T = %.1f lb-ft and S = %d rpm \n",T_d,S_d);
+
+printf(" \n e: At point w:\n T = %.1f lb-ft and S = %d rpm \n",T_e,S_e);
+
+printf(" \n f: P = %.4f hp = %.1f W \n ",P_d,P_d_watt);
+
+printf(" \n g: P = %.4f hp = %.f W \n",P_f,P_f_watt );
+
+printf(" \n h: A = %d W and B = %d W ", A , B );
diff --git a/1092/CH11/EX11.2/Example11_2.sce b/1092/CH11/EX11.2/Example11_2.sce new file mode 100755 index 000000000..29af191d2 --- /dev/null +++ b/1092/CH11/EX11.2/Example11_2.sce @@ -0,0 +1,28 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 11: SPECIALIZED DYNAMOS
+// Example 11-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// VR stepper motor
+n = 3 ; // Number of stacks or phases
+P_a = 16 ; // Number of rotor teeth (subscript a indicates case a)
+// PM stepper
+P_b = 24 ; // Number of poles (subscript b indicates case b)
+
+// Calculations
+// case a
+alpha_a = 360 / (n*P_a); // Stepping angle in degrees per step
+
+alpha_b = 360 / (n*P_b); // Stepping angle in degrees per step
+
+// Display the results
+disp("Example 11-2 Solution : ");
+printf(" \n a: alpha α = %.1f degrees/step \n ", alpha_a );
+
+printf(" \n b: alpha α = %.1f degrees/step \n ", alpha_b );
diff --git a/1092/CH11/EX11.3/Example11_3.sce b/1092/CH11/EX11.3/Example11_3.sce new file mode 100755 index 000000000..4be777467 --- /dev/null +++ b/1092/CH11/EX11.3/Example11_3.sce @@ -0,0 +1,21 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 11: SPECIALIZED DYNAMOS
+// Example 11-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Hybrid stepping motor
+P = 50 ; // Number of rotor teeth
+
+// Calculation
+
+alpha = 90 / P ; // Stepping angle in degrees
+
+// Display the result
+disp("Example 11-3 Solution : ");
+printf(" \n alpha α = %.1f degrees ", alpha );
diff --git a/1092/CH11/EX11.4/Example11_4.sce b/1092/CH11/EX11.4/Example11_4.sce new file mode 100755 index 000000000..b1dee2f5b --- /dev/null +++ b/1092/CH11/EX11.4/Example11_4.sce @@ -0,0 +1,20 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 11: SPECIALIZED DYNAMOS
+// Example 11-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+tou = 0.1 ; // Pole pitch of a double-sided primary LIM in meter
+f = 60 ; // Frequency applied to the primary LIM in Hz
+
+// Calculation
+v_s = 2 * f * tou ; // Synchronous velocity in meter/second
+
+// Display the result
+disp("Example 11-4 Solution : ");
+printf(" \n Synchronous velocity : \n v_s = %d m/s ", v_s );
diff --git a/1092/CH11/EX11.5/Example11_5.sce b/1092/CH11/EX11.5/Example11_5.sce new file mode 100755 index 000000000..b5e98a58f --- /dev/null +++ b/1092/CH11/EX11.5/Example11_5.sce @@ -0,0 +1,20 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 11: SPECIALIZED DYNAMOS
+// Example 11-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+v_s = 12 ; // Synchronous velocity in meter/second
+v = 10 ; // Secondary sheet in Ex.11-4 moves at a linear velocity in m/s
+
+// Calculation
+s = (v_s - v)/v_s ; // Slip of the DSLIM
+
+// Display the result
+disp("Example 11-5 Solution : ");disp("From Eq.(11-5)")
+printf(" \n Slip of the DSLIM : \n s = %.3f ",s );
diff --git a/1092/CH12/EX12.1/Example12_1.sce b/1092/CH12/EX12.1/Example12_1.sce new file mode 100755 index 000000000..4d58df864 --- /dev/null +++ b/1092/CH12/EX12.1/Example12_1.sce @@ -0,0 +1,50 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 10000 ; // Power rating of the shunt generator in W
+V = 230 ;// Voltage rating of the shunt generator in volt
+S = 1750 ; // Speed in rpm of the shunt generator
+// Shunt generator was made to run as a motor
+V_a = 245 ; // Voltage across armature in volt
+I_a = 2 ; // Armature current in A
+R_f = 230 ; // Field resistance in ohm
+R_a = 0.2 ; // Armature resistance
+
+// Calculations
+// case a
+Rotational_losses = (V_a * I_a) - (I_a^2 * R_a); // Rotational losses in W at full load
+
+// case b
+V_t = V ;
+// At rated load
+I_L = P / V_t ; // Line current in A
+I_f = V / R_f ; // Field current in A
+Ia = I_f + I_L ; // Armature current in A
+
+armature_loss = (Ia^2 * R_a); // Full-load armature loss in W
+V_f = V ; // Field voltage in volt
+field_loss = V_f * I_f; // Full-load field loss in W
+
+// case c
+//
+eta = P / ( P + Rotational_losses + (armature_loss+field_loss) ) * 100 ;
+
+// Display the results
+disp("Example 12-1 Solution : ");
+
+printf(" \n a: Rotational losses at full load = %.1f W \n",Rotational_losses);
+
+printf(" \n b: At the rated load,\n I_L = %.1f A\n I_a = %.1f A\n",I_L,Ia);
+printf(" \n Full-load armature loss :\n (I_a^2)*R_a = %.f W \n",armature_loss);
+printf(" \n Full-load field loss :\n V_f*I_f = %.f W \n",field_loss);
+
+printf(" \n c: Efficiency of the generator at rated load(full-load in this Ex.) : ");
+printf(" \n η = %.1f percent ",eta);
diff --git a/1092/CH12/EX12.10/Example12_10.sce b/1092/CH12/EX12.10/Example12_10.sce new file mode 100755 index 000000000..bb2d880a4 --- /dev/null +++ b/1092/CH12/EX12.10/Example12_10.sce @@ -0,0 +1,76 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 125 ; // Voltage rating of genrator in volt
+P_o = 12500 ; // Power rating of genrator in W
+P_hp = 20 ; // Power rating of motor in hp
+R_a = 0.1 ; // Armture resistance in ohm
+R_f = 62.5 ; // Field resistance in ohm
+P_var = 1040 ; // Rated variable electric loss in W
+
+// Calculations
+// case a
+P_in = P_hp * 746 ; // Power input to generator in W
+eta = P_o / P_in * 100 ; // Efficiency
+
+// case b
+V_f = V ; // Voltage across shunt field wdg in volt
+P_sh_loss = (V_f)^2 / R_f ; // Shunt field loss in W
+
+// case c
+V_L = V ;
+I_L = P_o / V_L ; // Line current in A
+I_f = V_f / R_f ; // Field current in A
+I_a = I_L + I_f ; // Armature current in A
+E_g = V_L + I_a*R_a ; // Generated EMF in volt
+
+P_d1 = E_g * I_a ; // Generated electric power in W
+P_f = V_f * I_f ;
+P_d2 = P_o + P_var + P_f ; // Generated electric power in W
+
+// case d
+P_d = P_d1;
+P_r = P_in - P_d ; // Rotational power losses in W
+
+// case e
+P_k = P_r + V_f*I_f ; // Constant losses in W
+Ia = sqrt(P_k/R_a); // Armature current in A for max.efficiency
+
+// case f
+I_a_rated = I_a ; // Rated armature current in A
+LF = Ia / I_a ; // Load fraction
+
+// case g
+rated_output = 12500 ; // Rated output in kW
+// Maximum efficiency
+eta_max = ( LF * rated_output ) / ( ( LF * rated_output ) + (2*P_k) ) * 100 ;
+
+// Display the results
+disp("Example 12-10 Solution : ");
+
+printf(" \n a: Efficiency :\n η = %f percent ≃ %.1f percent \n ",eta,eta);
+
+printf(" \n b: Shunt field loss :\n (V_f)^2/R_f = %d W \n ",P_sh_loss);
+
+printf(" \n c: Line current : I_L = %d A \n\n Field current : I_f = %d A",I_L,I_f);
+printf(" \n\n Armature current : I_a = %d A ",I_a);
+printf(" \n\n Generated EMF : E_g = %.1f V ",E_g);
+printf(" \n\n Generated electric power : ");
+printf(" \n 1. P_d = %d W \n\n 2. P_d = %d W \n ",P_d1,P_d2);
+
+printf(" \n d: Rotational power losses :\n P_r = %f W ≃ %.f W \n",P_r,P_r);
+
+printf(" \n e: Constant losses : P_k = %f W ≃ %.f W \n ", P_k ,P_k);
+printf(" \n Armature current for max.efficiency : I_a = %.1f A \n ",Ia);
+
+printf(" \n f: Load fraction : L.F. = %.2f \n ",LF);
+
+printf(" \n g: Maximum efficiency : η = %f percent ≃ %.2f percent",eta_max,eta_max);
diff --git a/1092/CH12/EX12.11/Example12_11.sce b/1092/CH12/EX12.11/Example12_11.sce new file mode 100755 index 000000000..1f77df453 --- /dev/null +++ b/1092/CH12/EX12.11/Example12_11.sce @@ -0,0 +1,59 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data (from Ex.12-10)
+V = 125 ; // Voltage rating of genrator in volt
+P_o = 12500 ; // Power rating of genrator in W
+P_hp = 20 ; // Power rating of motor in hp
+R_a = 0.1 ; // Armture resistance in ohm
+R_f = 62.5 ; // Field resistance in ohm
+P_var = 1040 ; // Rated variable electric loss in W
+
+// Calculated data from Ex.12-10
+P_k = 1380 ; // Constant losses in W
+
+// Calculations
+// Efficiency of the dc shunt generator
+// η = (output*L.F) / ( (output*L.F) + P_k + (L.F)^2 * P_a_rated ) * 100
+output = P_o ;
+P_a_rated = P_var ;
+
+// case a
+LF1 = 25*(1/100); // At 25 % rated output
+// Efficiency of the dc shunt generator at 25 % rated output
+eta_1 = (output*LF1) / ( (output*LF1) + P_k + (LF1)^2 * P_a_rated ) * 100 ;
+
+// case b
+LF2 = 50*(1/100); // At 50 % rated output
+// Efficiency of the dc shunt generator at 50 % rated output
+eta_2 = (output*LF2) / ( (output*LF2) + P_k + (LF2)^2 * P_a_rated ) * 100 ;
+
+// case c
+LF3 = 75*(1/100); // At 75 % rated output
+// Efficiency of the dc shunt generator at 75 % rated output
+eta_3 = (output*LF3) / ( (output*LF3) + P_k + (LF3)^2 * P_a_rated ) * 100 ;
+
+// case d
+LF4 = 125*(1/100); // At 125 % rated output
+// Efficiency of the dc shunt generator at 125 % rated output
+eta_4 = (output*LF4) / ( (output*LF4) + P_k + (LF4)^2 * P_a_rated ) * 100 ;
+
+
+// Display the results
+disp("Example 12-11 Solution : ");
+
+printf(" \n a: η at %.2f rated output = %.2f percent \n ",LF1,eta_1);
+
+printf(" \n b: η at %.2f rated output = %.2f percent \n ",LF2,eta_2);
+printf(" \n Please note: Calculation error for case b: η in the textbook.\n");
+
+printf(" \n c: η at %.2f rated output = %.2f percent \n ",LF3,eta_3);
+
+printf(" \n d: η at %.2f rated output = %.2f percent \n ",LF4,eta_4);
diff --git a/1092/CH12/EX12.12/Example12_12.sce b/1092/CH12/EX12.12/Example12_12.sce new file mode 100755 index 000000000..43c9ef617 --- /dev/null +++ b/1092/CH12/EX12.12/Example12_12.sce @@ -0,0 +1,101 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase Y-connected alternator
+kVA = 100 ; // kVA rating of the alternator
+V = 1100 ; // Rated voltage of the alternator in volt
+I_a_nl = 8 ; // No-load armature current in A
+P_in_nl = 6000 ; // No-load Power input to the armature in W
+V_oc = 1350 ; // Open-ckt line voltage in volt
+I_f = 18 ; // Field current in A
+V_f = 125 ; // voltage across field winding in volt
+
+// Calculations
+// From Ex.6-4,
+R_a = 0.45 ; // Armature resistance in ohm/phase
+I_a_rated = 52.5 ; // Rated armature current in A/phase
+
+// case a
+P_r = P_in_nl - 3 * (I_a_nl)^2 * R_a ; // Rotational loss of synchronous dynamo in W
+
+// case b
+P_f = V_f*I_f ; // Field copper loss in W
+
+// case c
+P_k = P_r + P_f ; // Fixed losses in W at rated synchronous speed
+Pk = P_k / 1000 ; // Fixed losses in kW at rated synchronous speed
+
+// case d
+P_cu = 3 * (I_a_rated)^2 * R_a ; // Rated electric armature cu-loss in W
+P_cu_kW = P_cu / 1000 ; // Rated electric armature cu-loss in kW
+
+LF1 = 1/4 ; // Load fraction
+LF2 = 1/2 ; // Load fraction
+LF3 = 3/4 ; // Load fraction
+P_cu_LF1 = P_cu * (LF1)^2 ; // Electric armature cu-loss in W at 1/4 load
+P_cu_LF2 = P_cu * (LF2)^2 ; // Electric armature cu-loss in W at 1/2 load
+P_cu_LF3 = P_cu * (LF3)^2 ; // Electric armature cu-loss in W at 3/4 load
+
+P_cu_LF1_kW = P_cu_LF1 / 1000 ; // Electric armature cu-loss in kW at 1/4 load
+P_cu_LF2_kW = P_cu_LF2 / 1000 ; // Electric armature cu-loss in kW at 1/2 load
+P_cu_LF3_kW = P_cu_LF3 / 1000 ; // Electric armature cu-loss in kW at 3/4 load
+
+
+// case e
+PF = 0.9 ; // Power factor lagging
+// Efficiency
+// η = LF(rated kVA)*PF / ( LF(rated kVA)*PF + P_k + P_cu ) * 100
+eta_1 = (LF1 * kVA * PF) / ( (LF1 * kVA * PF) + Pk + P_cu_LF1_kW ) * 100 ;// Efficiency at 1/4 load
+eta_2 = (LF2 * kVA * PF) / ( (LF2 * kVA * PF) + Pk + P_cu_LF2_kW ) * 100 ;// Efficiency at 1/2 load
+eta_3 = (LF3 * kVA * PF) / ( (LF3 * kVA * PF) + Pk + P_cu_LF3_kW ) * 100 ;// Efficiency at 3/4 load
+eta_fl = (kVA * PF) / ( (kVA * PF) + Pk + P_cu_kW ) * 100 ;// Efficiency at full load
+
+// case f
+Ia = sqrt(P_k/(3*R_a)); // Armature current in A for max.efficiency at 0.9 PF lagging
+LF = Ia / I_a_rated ; // Load fraction for max.efficiency
+// at max.efficiency P_cu = P_k
+eta_max = (LF * kVA * PF) / ( (LF * kVA * PF) + 2*Pk ) * 100 ;// Max Efficiency 0.9 PF lagging
+
+// case g
+P_o = kVA*PF ; // Output power at 0.9 PF lagging
+I_a = I_a_rated ;
+P_d = P_o + (3*(I_a)^2*R_a/1000) + (V_f*I_f/1000) ; // Armature power developed in kW at 0.9 PF lagging at full-load
+
+// Display the results
+disp("Example 12-12 Solution : ");
+
+printf(" \n From Ex.6-4,\n R_a = %.2f Ω/phase",R_a);
+printf(" \n I_a(rated) = %.1f A \n ",I_a_rated);
+
+printf(" \n a: Rotational loss of synchronous dynamo :\n P_r = %.f W \n",P_r);
+
+printf(" \n b: Field copper loss :\n P_f = %d W \n ",P_f);
+
+printf(" \n c: Fixed losses at rated synchronous speed :\n P_k = %.f W\n",P_k);
+
+printf(" \n d: P_cu at rated load = %.f W\n P_cu ,",P_cu);
+printf(" \n at %.2f rated load = %.1f W",LF1 , P_cu_LF1);
+printf(" \n at %.2f rated load = %.1f W",LF2 , P_cu_LF2);
+printf(" \n at %.2f rated load = %.1f W \n",LF3 , P_cu_LF3);
+
+
+printf(" \n e: Efficiency :\n η at %.2f load = %.1f percent",LF1,eta_1);
+printf(" \n η at %.2f load = %.1f percent",LF2,eta_2);
+printf(" \n η at %.2f load = %.1f percent",LF3,eta_3);
+printf(" \n η at full-load = %.1f percent \n",eta_fl);
+
+printf(" \n f: Armature current for max.efficiency at 0.9 PF lagging :");
+printf(" \n I_a(max) = %f A ≃ %.1f A\n",Ia,Ia);
+printf(" \n L.F. = %.2f \n",LF);
+printf(" \n Maximum efficiency :\n η_max = %.1f percent \n ",eta_max);
+
+printf(" \n g: Armature power developed at 0.9 PF lagging at full-load :");
+printf(" \n P_d = %.2f kW ",P_d);
diff --git a/1092/CH12/EX12.13/Example12_13.sce b/1092/CH12/EX12.13/Example12_13.sce new file mode 100755 index 000000000..f95985e73 --- /dev/null +++ b/1092/CH12/EX12.13/Example12_13.sce @@ -0,0 +1,107 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-13
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase Y-connected alternator
+kVA = 1000 ; // kVA rating of the alternator
+V = 2300 ; // Rated voltage of the alternator in volt
+
+// DC MOTOR
+P_hp = 100 ; // Power rating of the dc motor in hp
+V_motor = 240 ; // Rated voltage of the motor in volt
+// 4-step efficiency/regulation test
+// Test 1
+P_1 = 7.5 ; // motor output in kW
+
+// Test 2
+P_2 = 16 ; // motor output in kW
+VfIf = 14 ; // Field losses in kW
+P_f = VfIf ; // Field losses in kW
+
+// Test 3
+P_3 = 64.2 ; // motor output in kW
+I_sc = 251 ; // Short ckt current in A
+
+// Test 4
+V_L = 1443 ; // Line voltage in volt
+
+// Calculations
+// case a
+P_r = P_2 ; // Rotational losses in kW From test 2
+
+// case b
+P_cu = P_3 - P_1 ; // Full-load armature copper loss in kW
+
+// case c
+E_gL = V_L ; // Generated line voltage in volt
+Z_s = (E_gL/sqrt(3)) / I_sc ; // Synchronous impedance of the armature in ohm
+
+// case d
+R_a = 0.3 ; // Armature resistance in ohm
+X_s = sqrt( (Z_s)^2 - (R_a)^2 ); // Synchronous reactance of the armature in ohm
+
+// case e
+cos_theta = 0.8 ; // PF lagging
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+V_p = V / sqrt(3); // Phase voltage in volt
+
+// Generated voltage per phase in volt
+I_a = I_sc ; // Armature current in A
+
+E_gp = (V_p*cos_theta + I_a*R_a) + %i*(V_p*sin_theta + I_a*X_s);
+E_gp_m = abs(E_gp);//E_gp_m=magnitude of E_gp in volt
+E_gp_a = atan(imag(E_gp) /real(E_gp))*180/%pi;//E_gp_a=phase angle of E_gp in degrees
+
+V_nl = E_gp_m ; // No-load voltage in volt
+V_fl = V_p ; // Full-load voltage in volt
+
+VR = (V_nl - V_fl)/V_fl * 100 ; // Alternator voltage regulation
+
+// case f
+PF = 0.8 ; // lagging PF
+LF = 1 ; // load fraction
+eta_rated = (LF*kVA*PF)/( (LF*kVA*PF) + (P_f + P_r) + P_cu ) * 100 ; // Efficiency at 0.8 lagging PF
+
+// case g
+P_k = (P_f + P_r) ; // Constant losses in kW
+L_F = sqrt(P_k/P_cu); // Load fraction for max.efficiency
+// at max.efficiency P_k = P_cu
+eta_max = (L_F*kVA*PF)/( (L_F*kVA*PF) + 2*P_k ) * 100 ; // Max.Efficiency at 0.8 lagging PF
+
+
+// case h
+P_o = kVA ; // Output power in kVA
+P_d = P_o +(3*(I_a)^2*R_a/1000) + (VfIf) ; // Armature power developed in kW at unity PF at rated-load
+
+// Display the results
+disp("Example 12-13 Solution : ");
+
+printf(" \n a: From Test 2, Rotational losses :\n P_r = %d kW \n",P_r);
+
+printf(" \n b: Full-load armature copper loss :\n P_cu = %.1f kW \n",P_cu);
+
+printf(" \n c: Synchronous impedance of the armature :\n Z_s = %f Ω ≃ %.2f Ω \n",Z_s,Z_s);
+
+printf(" \n d: Synchronous reactance of the armature :\n jX_s = %f Ω ≃ %.2f Ω \n",X_s,X_s);
+
+printf(" \n e: E_gp = ");disp(E_gp);
+printf(" \n E_gp = %.f <%.1f V\n",E_gp_m,E_gp_a);
+printf(" \n Alternator voltage regulation :\n VR = %.2f percent \n",VR);
+
+printf(" \n Obtained VR value through scilab calculation is slightly different from textbook");
+printf(" \n because of non-approximation of Z_s,X_s and E_gp while calculating in scilab.\n");
+
+printf(" \n f: Alternator efficiency at 0.8 lagging PF :\n η_rated = %.1f percent\n",eta_rated);
+
+printf(" \n g: L.F = %.4f\n",L_F);
+printf(" \n Max.Efficiency at 0.8 lagging PF :\ η_max = %.2f percent \n",eta_max );
+
+printf(" \n h: Power developed by the alternator armature at rated load,unity PF :");
+printf(" \n P_d = %.f kW",P_d);
diff --git a/1092/CH12/EX12.14/Example12_14.sce b/1092/CH12/EX12.14/Example12_14.sce new file mode 100755 index 000000000..99ce627c5 --- /dev/null +++ b/1092/CH12/EX12.14/Example12_14.sce @@ -0,0 +1,110 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-14
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 4 ;// Number of poles in Induction motor
+f = 60 ; // Frequency in Hz
+V = 220 ; // Rated voltage of IM in volt
+hp_IM = 5 ; // Power rating of IM in hp
+PF = 0.9 ; // Power factor
+I_L = 16 ; // Line current in A
+S = 1750 ; // Speed of IM in rpm
+
+// No-load test data
+I_nl = 6.5 ; // No-load line current in A
+V_nl = 220 ; // No-load line voltage in volt
+P_nl = 300 ; // No-load power reading in W
+
+// Blocked rotor test
+I_br = 16 ; // Blocked rotor line current in A
+V_br = 50 ; // Blocked rotor voltage in volt
+P_br = 800 ; // Blocked rotor power reading in W
+
+// Calculations
+// case a
+P_cu = P_br ; // Full-load equivalent cu-loss
+I_1 = I_br ; // Primary current in A
+R_e1 = (P_cu) / (3/2 * (I_1)^2 ); // Equivalent total resistance of IM in ohm
+
+// case b
+P_in = P_nl ; // Input power to IM
+I1 = I_nl ; // Input current in A
+P_r = P_in - (3/2 * (I1)^2 * R_e1); // Rotational losses in W
+
+// case c
+LF1 = 1/4 ; // Load fraction
+LF2 = 1/2 ; // Load fraction
+LF3 = 3/4 ; // Load fraction
+LF4 = 5/4 ; // Load fraction
+P_cu_LF1 = (LF1)^2 * P_cu ; // Equivalent copper loss at 1/4 rated-load
+P_cu_LF2 = (LF2)^2 * P_cu ; // Equivalent copper loss at 1/2 rated-load
+P_cu_LF3 = (LF3)^2 * P_cu ; // Equivalent copper loss at 3/4 rated-load
+P_cu_LF4 = (LF4)^2 * P_cu ; // Equivalent copper loss at 5/4 rated-load
+
+// case d
+Full_load_input = sqrt(3)*V*I_L*PF ;
+
+// Efficiency
+// Efficiency at 1/4 rated load
+eta_LF1 = ( Full_load_input*LF1 - (P_r + P_cu_LF1) ) / (Full_load_input*LF1) * 100 ;
+
+// Efficiency at 1/2 rated load
+eta_LF2 = ( Full_load_input*LF2 - (P_r + P_cu_LF2) ) / (Full_load_input*LF2) * 100 ;
+
+// Efficiency at 3/4 rated load
+eta_LF3 = ( Full_load_input*LF3 - (P_r + P_cu_LF3) ) / (Full_load_input*LF3) * 100 ;
+
+// Efficiency at rated load
+eta_rated = ( Full_load_input - (P_r + P_cu) ) / (Full_load_input) * 100 ;
+
+// Efficiency at 5/4 rated load
+eta_LF4 = ( Full_load_input*LF4 - (P_r + P_cu_LF4) ) / (Full_load_input*LF4) * 100 ;
+
+// case e
+// since eta is calculated in percent divide it by 100 for hp calculations
+P_o_LF1 = (Full_load_input*LF1*eta_LF1/100)/746 ; // Output hp at 1/4 rated load
+P_o_LF2 = (Full_load_input*LF2*eta_LF2/100)/746 ; // Output hp at 1/2 rated load
+P_o_LF3 = (Full_load_input*LF3*eta_LF3/100)/746 ; // Output hp at 3/4 rated load
+P_o = (Full_load_input*eta_rated/100)/746 ; // Output hp at 1/4 rated load
+P_o_LF4 = (Full_load_input*LF4*eta_LF4/100)/746 ; // Output hp at 5/4 rated load
+
+// case f
+hp = P_o ; // Rated output horsepower
+T_o = (P_o*5252)/S ; // Outpue torque at full-load in lb-ft
+T_o_Nm = T_o * 1.356 ; // Outpue torque at full-load in N-m
+
+// Display the results
+disp("Example 12-14 Solution : ");
+
+printf(" \n a: Equivalent total resistance of IM :\n R_e1 = %.3f Ω \n",R_e1);
+
+printf(" \n b: Rotational losses :\n P_r = %.f W \n ",P_r);
+
+printf(" \n c: At full-load, P_cu = %d W \n",P_cu);
+printf(" \n P_cu at %.2f rated load = %d W",LF1,P_cu_LF1)
+printf(" \n P_cu at %.2f rated load = %d W",LF2,P_cu_LF2)
+printf(" \n P_cu at %.2f rated load = %d W",LF3,P_cu_LF3)
+printf(" \n P_cu at %.2f rated load = %d W \n",LF4,P_cu_LF4)
+
+printf(" \n d: Full-load input = %.f W \n",Full_load_input);
+printf(" \n Efficiency :\n η at %.2f rated load = %.1f percent \n",LF1,eta_LF1);
+printf(" \n η at %.2f rated load = %.1f percent \n",LF2,eta_LF2);
+printf(" \n η at %.2f rated load = %.1f percent \n",LF3,eta_LF3);
+printf(" \n η at rated load = %.1f percent \n",eta_rated);
+printf(" \n η at %.2f rated load = %.1f percent \n",LF4,eta_LF4);
+
+printf(" \n e: Output horsepower :\n P_o at %.2f rated load = %.3f hp \n",LF1,P_o_LF1);
+printf(" \n P_o at %.2f rated load = %.3f hp \n",LF2,P_o_LF2);
+printf(" \n P_o at %.2f rated load = %.3f hp \n",LF3,P_o_LF3);
+printf(" \n P_o at rated load = %.3f hp \n",P_o);
+printf(" \n P_o at %.2f rated load = %.3f hp \n",LF4,P_o_LF4);
+
+printf(" \n f: Output torque at full-load :\n T_o = %.1f lb-ft",T_o);
+printf(" \n T_o = %.2f N-m",T_o_Nm);
diff --git a/1092/CH12/EX12.15/Example12_15.sce b/1092/CH12/EX12.15/Example12_15.sce new file mode 100755 index 000000000..52f5d76ff --- /dev/null +++ b/1092/CH12/EX12.15/Example12_15.sce @@ -0,0 +1,86 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-15
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data(from Ex.12-14)
+pole = 4 ;// Number of poles in Induction motor
+f = 60 ; // Frequency in Hz
+V = 220 ; // Rated voltage of IM in volt
+hp_IM = 5 ; // Power rating of IM in hp
+PF = 0.9 ; // Power factor
+I_L = 16 ; // Line current in A
+S_r = 1750 ; // Speed of IM in rpm
+
+// No-load test data
+I_nl = 6.5 ; // No-load line current in A
+V_nl = 220 ; // No-load line voltage in volt
+P_nl = 300 ; // No-load power reading in W
+
+// Blocked rotor test
+I_br = 16 ; // Blocked rotor line current in A
+V_br = 50 ; // Blocked rotor voltage in volt
+P_br = 800 ; // Blocked rotor power reading in W
+R_dc = 1 ; // dc resistance in ohm between lines
+
+// given data from ex.12-15
+V = 220 ; // voltage rating in volt
+P_input = 5500 ; // power drawn in W
+
+// Calculations
+// Preliminary calculations
+R_e1 = 1.25*R_dc ; // Equivalent total resistance of IM in ohm
+P_in = P_nl ; // Input power to IM in W
+I1 = I_nl ; // Input current in A
+P_r = P_in - (3/2 * (I1)^2 * R_e1); // Rotational losses in W
+
+I_1 = I_L ;
+SCL = (3/2 * (I_1)^2 * R_e1) ; // Stator Copper Loss in W at full-load
+SPI = P_input ; // Stator Power Input in W
+RPI = SPI - SCL ; // Rotor Power Input in W
+
+S = (120*f/pole); // Speed of synchronous magnetic field in rpm
+s = (S-S_r)/S ; // Slip
+
+RPD = RPI*(1-s); // Rotor Power Developed in W
+RPO = RPD - P_r ; // Rotor Power Output in W
+
+// case a
+P_o = RPO ;
+eta_fl = (P_o / P_input)*100 ; // Full-load efficiency
+
+// case b
+hp = P_o / 746 ; // Output horsepower
+T_o = (hp*5252)/S_r ; // Output torque in lb-ft
+T_o_Nm = T_o * 1.356 ; // Output torque in N-m
+
+// Display the results
+disp("Example 12-15 Solution : ");
+
+printf(" \n Preliminary calculations :");
+printf(" \n R_e1 = %.2f Ω \n",R_e1);
+printf(" \n P_r = %.1f W \n ",P_r);
+printf(" \n SCL(fl) = %d W \n ",SCL);
+printf(" \n RPI(fl) = %d W \n ",RPI);
+printf(" \n RPD(fl) = %f W ≃ %.1f W \n ",RPD,RPD);
+printf(" \n RPO(fl) = %f W ≃ %.f W \n ",RPO,RPO);
+
+printf(" \n a: Full-load efficiency :\n η_fl = %.1f percent \n",eta_fl);
+
+printf(" \n b: Output horsepower :\n hp = %.2f hp at full-load \n",hp);
+printf(" \n Output torque at full-load :\n T_o = %f lb-ft ≃ %.1f lb-ft",T_o,T_o);
+printf(" \n T_o = %f lb-ft ≃ %.2f N-m \n ",T_o_Nm,T_o_Nm);
+
+printf(" \n c: Comparision of results");
+printf(" \n ________________________________________________________________");
+printf(" \n \t\t\t\t\t Ex.12-14\tEx.12-15");
+printf(" \n ________________________________________________________________");
+printf(" \n \t η_fl(percent) \t\t\t 82.4 \t\t %.1f ",eta_fl);
+printf(" \n \t Rated output(hp) \t\t 6.06 \t\t %.2f ",hp);
+printf(" \n \t Rated output torque(lb-ft) \t 18.2 \t\t %.1f ",T_o);
+printf(" \n ________________________________________________________________");
diff --git a/1092/CH12/EX12.16/Example12_16.sce b/1092/CH12/EX12.16/Example12_16.sce new file mode 100755 index 000000000..ad3581707 --- /dev/null +++ b/1092/CH12/EX12.16/Example12_16.sce @@ -0,0 +1,135 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-16
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// code letter = J
+P = 6 ; // Number of poles
+S_r = 1176 ; // rotor speed in rpm
+V = 220 ; // Rated voltage of SCIM in volt
+f = 60 ; // Frequency in Hz
+hp_SCIM = 7.5 ; // Power rating of SCIM in hp
+
+R_ap = 0.3 ; // armature resistance in ohm/phase
+R_r = 0.144 ; // rotor resistance in ohm/phase
+jX_m = 13.5 ; // reactance in ohm/phase
+jX_s = 0.5 ; // synchronous reactance in ohm/phase
+jX_lr = 0.2 ; // Locked rotor reactance in ohm/phase
+P_r = 300 ; // Rotational losses in W
+
+disp("Example 12-16 : ");
+// Calculations
+S = (120*f/P); // Speed of synchronous magnetic field in rpm
+// case a
+s = (S-S_r)/S ; // Slip
+
+R_r_by_s = R_r / s ;
+
+// case b
+printf(" \n From fig.12-11 , using the format method of mesh analysis,we may write");
+printf(" \n the array by inspection :\n ");
+printf(" \n_______________________________________________");
+printf(" \n \t I_1 \t I_2 \t\t V ");
+printf(" \n_______________________________________________");
+printf(" \n\t (0.3+j14) -(0+j13.5) \t(127+j0)");
+printf(" \n\t-(0+j13.5) (7.2+j13.7) \t 0");
+printf(" \n_______________________________________________\n");
+
+A = [ (0.3 + %i*14) -%i*13.5 ; (-%i*13.5) (7.2 + %i*13.7) ]; // Matrix containing above mesh eqns array
+delta = det(A); // Determinant of A
+
+// case b : Stator armature current I_p in A
+I_p = det( [ (127+%i*0) (-%i*13.5) ; 0 (7.2 + %i*13.7) ] ) / delta ;
+I_p_m = abs(I_p);//I_p_m=magnitude of I_p in A
+I_p_a = atan(imag(I_p) /real(I_p))*180/%pi;//I_p_a=phase angle of I_p in degrees
+I_1 = I_p ; // Stator armature current in A
+
+// case c : Rotor current I_r per phase in A
+I_r = det( [ (0.3 + %i*14) (127+%i*0) ; (-%i*13.5) 0 ] ) / delta ;
+I_r_m = abs(I_r);//I_r_m=magnitude of I_r in A
+I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees
+
+// case d
+theta = I_p_a ; // Motor PF angle in degrees
+cos_theta = cosd(theta); // Motor PF
+
+// case e
+I_p = I_p_m ; // Stator armature current in A
+V_p = V / sqrt(3); // Phase voltage in volt
+SPI = V_p * I_p * cos_theta ; // Stator Power Input in W
+
+// case f
+SCL = (I_p)^2 * R_ap ; // Stator Copper Loss in W
+
+// case g
+// Subscripts 1 and 2 for RPI indicates two methods of calculating RPI
+RPI_1 = SPI - SCL ; // Rotor Power Input in W
+RPI_2 = (I_r_m)^2 * (R_r/s); // Rotor Power Input in W
+RPI =RPI_1 ;
+
+// case h
+RCL = s*(RPI); // Rotor copper losses in W
+
+// case i
+// Subscripts 1 , 2 and 3 for RPD indicates three methods of calculating RPD
+RPD_1 = RPI - RCL ; // Rotor Power Developed in W
+RPD_2 = RPI * ( 1 - s ); // Rotor Power Developed in W
+RPD = RPD_1 ;
+
+// case j
+RPO = 3*RPD - P_r ; // Rotor Power Developed in W
+
+// case k
+P_to = RPO ; // Total rotor power in W
+T_o = (7.04*P_to)/S_r ; // Total 3-phase torque in lb-ft
+
+// case l
+hp = P_to / 746 ; // Output horsepower
+
+// case m
+P_in = 3*SPI ; // Input power to stator in W
+P_o = RPO ; // Output power in W
+eta = P_o / P_in * 100 ; // Motor efficiency at rated load
+
+// Display the results
+disp("Solution : ");
+printf(" \n a: s = %.2f \n R_r/s = %.1f Ω \n",s,R_r_by_s );
+
+printf(" \n Determinant Δ = ");disp(delta);
+
+printf(" \n b: Stator armature current :\n I_p in A = ");disp(I_1);
+printf(" \n I_p = I_1 = %.2f <%.2f A \n ",I_p_m , I_p_a );
+
+printf(" \n c: Rotor current per phase :\n I_r in A = ");disp(I_r);
+printf(" \n I_r = I_2 = %.3f <%.2f A \n ",I_r_m , I_r_a );
+
+printf(" \n d: Motor PF :\n cosӨ = %.4f \n",cos_theta);
+
+printf(" \n e: Stator Power Input :\n SPI = %d W \n",SPI);
+
+printf(" \n f: Stator Copper Loss :\n SCL = %.1f W \n",SCL);
+
+printf(" \n g: Rotor Power Input :\n RPI = %.1f W(method 1) ", RPI_1);
+printf(" \n RPI = %.1f W (method 2)\n",RPI_2);
+
+printf(" \n h: Rotor copper loss :\n RCL = %.1f W\n",RCL);
+
+printf(" \n i: Rotor Power Developed :\n RPD = %.1f W \n",RPD_1);
+
+printf(" \n RPD = %.1f W \n ",RPD_2);
+
+printf(" \n j: Total 3-phase rotor power:\n RPO = %f W \n",RPO);
+
+printf(" \n k: Total output torque developed :\n T_o = %.2f lb-ft\n",T_o);
+
+printf(" \n l: Output horsepower : \n hp = %.2f hp (rated 7.5 hp)\n",hp);
+
+printf(" \n m: Motor efficiency at rated load :\n η = %.2f percent \n",eta);
+
+printf(" \n n: See Fig.12-12");
diff --git a/1092/CH12/EX12.17/Example12_17.sce b/1092/CH12/EX12.17/Example12_17.sce new file mode 100755 index 000000000..837cf4e97 --- /dev/null +++ b/1092/CH12/EX12.17/Example12_17.sce @@ -0,0 +1,38 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-17
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// code letter = J of SCIM (Ex.12-16)
+
+// Calculations
+// case a
+// From Appendix A-3,Table 430-7(b),the starting kVA/hp (with rotor locked) is
+// less than 7.99,which, when substituted in the following equation, yields a
+// maximum starting current of :
+
+// subscript u for I_s indicates upper limit of starting current
+I_s_u = (7.99*(7.5*1000))/(sqrt(3)*220) ;
+
+// case b
+// The lower limit,code letter J,is 7.1 kVA/hp. Thus :
+
+// subscript l for I_s indicates lower limit of starting current
+I_s_l = (7.1*(7.5*1000))/(sqrt(3)*220) ;
+
+// Display the results
+disp("Example 12-17 Solution : ");
+
+printf(" \n a: From Appendix A-3,Table 430-7(b),the starting kVA/hp ");
+printf(" \n (with rotor locked) is less than 7.99,which, when substituted ");
+printf(" \n in the following equation, yields a maximum starting current of :");
+printf(" \n I_s = %.1f A \n",I_s_u);
+
+printf(" \n b: The lower limit,code letter J,is 7.1 kVA/hp.\n Thus :");
+printf(" \n I_s = %.1f A ",I_s_l );
diff --git a/1092/CH12/EX12.18/Example12_18.sce b/1092/CH12/EX12.18/Example12_18.sce new file mode 100755 index 000000000..0e0b3e7ad --- /dev/null +++ b/1092/CH12/EX12.18/Example12_18.sce @@ -0,0 +1,60 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-18
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data (Ex.12-16)
+// code letter = J
+P = 6 ; // Number of poles
+S_r = 1176 ; // rotor speed in rpm
+V = 220 ; // Rated voltage of SCIM in volt
+f = 60 ; // Frequency in Hz
+hp_SCIM = 7.5 ; // Power rating of SCIM in hp
+
+R_ap = 0.3 ; // armature resistance in ohm/phase
+R_r = 0.144 ; // rotor resistance in ohm/phase
+jX_m = 13.5 ; // reactance in ohm/phase
+jX_s = 0.5 ; // synchronous reactance in ohm/phase
+jX_lr = 0.2 ; // Locked rotor reactance in ohm/phase
+P_r = 300 ; // Rotational losses in W
+s = 1 ; // unity slip
+
+disp("Example 12-18 Solution : ");
+
+printf(" \n The ratio R_r/s = %.3f ohm, in fig.12-11 , using the format method ",R_r/s);
+printf(" \n of mesh analysis,we may write the array by inspection :\n ");
+printf(" \n_______________________________________________");
+printf(" \n \t I_1 \t I_2 \t\t V ");
+printf(" \n_______________________________________________");
+printf(" \n\t (0.3+j14) -(0+j13.5) \t(127+j0)");
+printf(" \n\t-(0+j13.5) (0.144+j13.7) \t 0");
+printf(" \n_______________________________________________\n");
+
+// Calculations
+
+A = [ (0.3 + %i*14) -%i*13.5 ; (-%i*13.5) (0.144 + %i*13.7) ]; // Matrix containing above mesh eqns array
+delta = det(A); // Determinant of A
+
+// case a : Starting stator current I_s per phase in A
+I_s = det( [ (127+%i*0) (-%i*13.5) ; 0 (0.144 + %i*13.7) ] ) / delta ;
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// case b : power factor of the motor at starting
+theta = I_s_a ; // Motor PF angle in degrees
+cos_theta = cosd(theta); // Motor PF
+
+// Display the results
+disp("Solution : ");
+printf(" \n a: Starting stator current of SCIM :\n I_s = I_1 = ");disp(I_s);
+printf(" \n I_s = I_1 = %.2f <%.2f A \n ",I_s_m , I_s_a );
+
+printf(" \n b: Power factor of the motor at starting :\n cosӨ = %.4f ≃ %.3f\n",cos_theta,cos_theta);
+
+printf(" \n Note : I_s = %.2f A calculated in Ex.12-18 falls between the limits",I_s_m);
+printf(" \n found in Ex.12-17. This verifies the mesh analysis technique.");
diff --git a/1092/CH12/EX12.19/Example12_19.sce b/1092/CH12/EX12.19/Example12_19.sce new file mode 100755 index 000000000..08d6517f9 --- /dev/null +++ b/1092/CH12/EX12.19/Example12_19.sce @@ -0,0 +1,146 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-19
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 220 ; // Rated voltage of SCIM in volt
+f = 60 ; // Frequency in Hz
+P = 4 ; // Number of poles
+PF = 0.85 ; // power factor of capacitor start IM
+// nameplate details
+hp_IM = 5 ; // power rating of IM in hp
+I_L = 28 ; // Rated line current in A
+S_r = 1620 ; // Rotor speed of IM in rpm
+
+// No-load test data
+I_nl = 6.4 ; // No-load line current in A
+V_nl = 220 ; // No-load line voltage in volt
+P_nl = 239 ; // No-load power reading in W
+s_nl = 0.01 ; // No-load slip
+
+// Blocked rotor test
+I_br = 62 ; // Blocked rotor line current in A
+V_br = 64 ; // Blocked rotor voltage in volt
+P_br = 1922 ; // Blocked rotor power reading in W
+s_br = 1 ; // blocked rotor slip(unity)
+
+// Calculations
+// case a
+R_e1s = P_br / (I_br^2); // Equivalent total resistance of IM in ohm
+
+// case b
+P_in = P_nl ; // Input power to IM in W
+I_1s = I_nl ; // Input current in A
+P_ro = P_in - ((I_1s)^2 * R_e1s); // Rotational losses in W
+
+// case c
+S = (120*f/P); // Speed of synchronous magnetic field in rpm
+S_fl = S_r ; // Full-load rotor speed of IM in rpm
+s_fl = (S - S_fl)/S ; // Full-load Slip
+
+LF1 = 1/4 ; // Load fraction
+LF2 = 1/2 ; // Load fraction
+LF3 = 3/4 ; // Load fraction
+LF4 = 5/4 ; // Load fraction
+
+s_LF1 = s_fl*LF1 ; // slip at 1/4 rated load
+s_LF2 = s_fl*LF2 ; // slip at 1/2 rated load
+s_LF3 = s_fl*LF3 ; // slip at 3/4 rated load
+s_LF4 = s_fl*LF4 ; // slip at 5/4 rated load
+
+// case d
+s_o = s_nl ; // No-load slip
+P_rs_LF1 = P_ro * (1 - s_LF1)/(1 - s_o); // Rotational losses in W at s_LF1
+P_rs_LF2 = P_ro * (1 - s_LF2)/(1 - s_o); // Rotational losses in W at s_LF2
+P_rs_LF3 = P_ro * (1 - s_LF3)/(1 - s_o); // Rotational losses in W at s_LF3
+P_rs_fl = P_ro * (1 - s_fl)/(1 - s_o); // Rotational losses in W at full-load slip
+P_rs_LF4 = P_ro * (1 - s_LF4)/(1 - s_o); // Rotational losses in W at s_LF4
+
+// case e
+I1s = I_L ; // Line current in A
+P_cu_fl = (I1s)^2*R_e1s ; // Equivalent copper loss at full-load slip
+P_cu_LF1 = (LF1)^2 * P_cu_fl ; // Equivalent copper loss at s_LF1
+P_cu_LF2 = (LF2)^2 * P_cu_fl ; // Equivalent copper loss at s_LF2
+P_cu_LF3 = (LF3)^2 * P_cu_fl ; // Equivalent copper loss at s_LF3
+P_cu_LF4 = (LF4)^2 * P_cu_fl ; // Equivalent copper loss at s_LF4
+
+// case f
+Input = V*I_L*PF ; // Input to single phase capacitor start IM
+
+// Efficiency at 1/4 rated load
+eta_LF1 = ( Input*LF1 - (P_rs_LF1 + P_cu_LF1) ) / (Input*LF1) * 100 ;
+
+// Efficiency at 1/2 rated load
+eta_LF2 = ( Input*LF2 - (P_rs_LF2 + P_cu_LF2) ) / (Input*LF2) * 100 ;
+
+// Efficiency at 3/4 rated load
+eta_LF3 = ( Input*LF3 - (P_rs_LF3 + P_cu_LF3) ) / (Input*LF3) * 100 ;
+
+// Efficiency at rated load
+eta_fl = ( Input - (P_rs_fl + P_cu_fl) ) / (Input) * 100 ;
+
+// Efficiency at 5/4 rated load
+eta_LF4 = ( Input*LF4 - (P_rs_LF4 + P_cu_LF4) ) / (Input*LF4) * 100 ;
+
+// case g
+// since eta is calculated in percent divide it by 100 for hp calculations
+P_o_LF1 = (Input*LF1*eta_LF1/100)/746 ; // Output hp at 1/4 rated load
+P_o_LF2 = (Input*LF2*eta_LF2/100)/746 ; // Output hp at 1/2 rated load
+P_o_LF3 = (Input*LF3*eta_LF3/100)/746 ; // Output hp at 3/4 rated load
+P_o = (Input*eta_fl/100)/746 ; // Output hp at 1/4 rated load
+P_o_LF4 = (Input*LF4*eta_LF4/100)/746 ; // Output hp at 5/4 rated load
+
+// case h
+hp = P_o ; // Rated output horsepower
+S_fl = S_r ; // Full-load rotor speed in rpm
+T_o = (P_o*5252)/S_fl ; // Outpue torque at full-load in lb-ft
+T_o_Nm = T_o * 1.356 ; // Outpue torque at full-load in N-m
+
+// Display the results
+disp("Example 12-19 Solution : ");
+
+printf(" \n a: Equivalent total resistance of IM :\n R_e1s = %.1f Ω \n",R_e1s);
+
+printf(" \n b: Rotational losses :\n P_ro = %.1f W \n ",P_ro);
+
+printf(" \n c: Slip at rated load : s = %.1f \n Slip,",s_fl);
+printf(" \n s at %.2f rated load = %.3f",LF1,s_LF1);
+printf(" \n s at %.2f rated load = %.3f",LF2,s_LF2);
+printf(" \n s at %.2f rated load = %.3f",LF3,s_LF3);
+printf(" \n s at %.2f rated load = %.3f \n ",LF4,s_LF4);
+
+printf(" \n d: Rotational losses :\n ");
+printf(" \n P_r at at %.2f rated load = %.1f W ",LF1,P_rs_LF1);
+printf(" \n P_r at at %.2f rated load = %.1f W ",LF2,P_rs_LF2);
+printf(" \n P_r at at %.2f rated load = %.1f W ",LF3,P_rs_LF3);
+printf(" \n P_r at at full load = %.1f W ",P_rs_fl);
+printf(" \n P_r at at %.2f rated load = %.1f W \n ",LF4,P_rs_LF4);
+
+printf(" \n e: At full-load, P_cu = %d W \n",P_cu_fl);
+printf(" \n P_cu at %.2f rated load = %.2f W",LF1,P_cu_LF1)
+printf(" \n P_cu at %.2f rated load = %.2f W",LF2,P_cu_LF2)
+printf(" \n P_cu at %.2f rated load = %.2f W",LF3,P_cu_LF3)
+printf(" \n P_cu at %.2f rated load = %.2f W \n",LF4,P_cu_LF4)
+
+printf(" \n f: Full-load input = %.f W \n",Input);
+printf(" \n Efficiency :\n η at %.2f rated load = %.1f percent \n",LF1,eta_LF1);
+printf(" \n η at %.2f rated load = %.1f percent \n",LF2,eta_LF2);
+printf(" \n η at %.2f rated load = %.1f percent \n",LF3,eta_LF3);
+printf(" \n η at rated load = η_fl = %.1f percent \n",eta_fl);
+printf(" \n η at %.2f rated load = %.1f percent \n",LF4,eta_LF4);
+printf(" \n Please note: Calculation error for η_fl in textbook.\n");
+
+printf(" \n g: Output horsepower :\n P_o at %.2f rated load = %.3f hp \n",LF1,P_o_LF1);
+printf(" \n P_o at %.2f rated load = %.3f hp \n",LF2,P_o_LF2);
+printf(" \n P_o at %.2f rated load = %.3f hp \n",LF3,P_o_LF3);
+printf(" \n P_o at rated load = %.3f hp \n",P_o);
+printf(" \n P_o at %.2f rated load = %.3f hp \n",LF4,P_o_LF4);
+
+printf(" \n h: Output torque at full-load :\n T_o = %.1f lb-ft",T_o);
+printf(" \n T_o = %.2f N-m ≃ %.1f N-m",T_o_Nm,T_o_Nm);
diff --git a/1092/CH12/EX12.2/Example12_2.sce b/1092/CH12/EX12.2/Example12_2.sce new file mode 100755 index 000000000..5820a56b3 --- /dev/null +++ b/1092/CH12/EX12.2/Example12_2.sce @@ -0,0 +1,56 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// data from Ex.12-1
+P = 10000 ; // Power rating of the shunt generator in W
+V = 230 ;// Voltage rating of the shunt generator in volt
+S = 1750 ; // Speed in rpm of the shunt generator
+
+// ( Solutions from Example 12-1 )
+Rotational_losses = 489.2 // Rotational losses at full load in W
+armature_loss = 396 ; // Full-load armature loss in W
+field_loss = 230 ; // Full-load field loss in W
+
+// case a
+x1 = (1/4); // Fraction of full-load
+// Subscript a for eta indicates case a
+eta_a = (P*x1) / ( (P*x1) + Rotational_losses + (armature_loss*(x1^2)+field_loss) ) * 100 ;
+
+// case b
+x2 = (1/2); // Fraction of full-load
+// Subscript b for eta indicates case b
+eta_b = (P*x2) / ( (P*x2) + Rotational_losses + (armature_loss*(x2^2)+field_loss) ) * 100 ;
+
+// case c
+x3 = (3/4); // Fraction of full-load
+// Subscript c for eta indicates case c
+eta_c = (P*x3) / ( (P*x3) + Rotational_losses + (armature_loss*(x3^2)+field_loss) ) * 100 ;
+
+// case d
+x4 = (5/4); // Fraction of full-load
+// Subscript d for eta indicates case d
+eta_d = (P*x4) / ( (P*x4) + Rotational_losses + (armature_loss*(x4^2)+field_loss) ) * 100 ;
+
+// Display the results
+disp("Example 12-2 Solution : ");
+
+printf(" \n If x is the fraction of full-load, then \n ");
+printf(" \n a: Efficiency of generator when x = %.2f ",x1 );
+printf(" \n η = %.1f percent \n ",eta_a);
+
+printf(" \n b: Efficiency of generator when x = %.2f ",x2 );
+printf(" \n η = %.1f percent \n ",eta_b);
+
+printf(" \n c: Efficiency of generator when x = %.2f ",x3 );
+printf(" \n η = %.1f percent \n ",eta_c);
+
+printf(" \n d: Efficiency of generator when x = %.2f ",x4 );
+printf(" \n η = %.1f percent \n ",eta_d);
diff --git a/1092/CH12/EX12.3/Example12_3.sce b/1092/CH12/EX12.3/Example12_3.sce new file mode 100755 index 000000000..dfcfc864a --- /dev/null +++ b/1092/CH12/EX12.3/Example12_3.sce @@ -0,0 +1,42 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 240 ; // Voltage rating of the dc shunt motor in volt
+P_hp = 25 ; // Power rating of the dc shunt motor in hp
+S = 1800 ; // Speed in rpm of the shunt generator
+I_L = 89 ; // Full-load line current
+R_a = 0.05 ; // Armature resistance in ohm
+R_f = 120 ; // Field resistance in ohm
+
+// Calculations
+// case a
+V_f = V ; // Field voltage in volt
+I_f = V_f / R_f ; // Field current in A
+I_a = I_L - I_f ; // Armature current in A
+V_a = V ;
+E_c = V_a - I_a*R_a ; // Armature voltage to be applied to the motor when motor
+// is run light at 1800 rpm during stray power test
+
+// case b
+Ia = 4.2 ; // Armature current in A produced by E_c
+Va = E_c ; // Armature voltage in volt
+P_r = Va*Ia ; // Stray power in W ,when E_c produces I_a = 4.2 A at speed of 1800 rpm
+
+// Display the results
+disp("Example 12-3 Solution : ");
+
+printf(" \n a: Field current :\n I_f = %d A \n ",I_f );
+printf(" \n Armature current :\n I_a = %d A \n ",I_a );
+printf(" \n Armature voltage to be applied to the motor when motor is run");
+printf(" \n light at %d rpm during stray power test :\n ",S );
+printf(" \n E_c = %.2f V \n ",E_c );
+
+printf(" \n b: Stray power :\n P_r = %.1f W ",P_r );
diff --git a/1092/CH12/EX12.4/Example12_4.sce b/1092/CH12/EX12.4/Example12_4.sce new file mode 100755 index 000000000..140b633ce --- /dev/null +++ b/1092/CH12/EX12.4/Example12_4.sce @@ -0,0 +1,136 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 600 ; // Voltage rating of the compound motor in volt
+P_hp = 150 ; // Power rating of the compound motor in hp
+I_L = 205 ; // Full-load rated line current in A
+S = 1500 ; // Full-load Speed in rpm of the compound generator
+R_sh = 300 ; // Shunt field resistance in ohm
+R_a = 0.05 ; // Armature resistance in ohm
+R_s = 0.1 ; // Series field resistance in ohm
+V_a = 570 ; // Applied voltage in volt
+I_a = 6 ; // Armature current in A
+S_o = 1800 ; // No-load Speed in rpm of the compound generator
+
+// Calculations
+// case a
+Rot_losses = V_a*I_a ; // Rotational losses in W
+// If x is fraction of full-load
+x1 = (1/4);
+S_1 = S_o - 300*x1 ; // Speed at 1/4 load
+Rot_losses_S_1 = (S_1/S)*Rot_losses ; // Rotational losses in W at speed S_1
+
+x2 = (1/2);
+S_2 = S_o - 300*x2 ; // Speed at 1/2 load
+Rot_losses_S_2 = (S_2/S)*Rot_losses ; // Rotational losses in W at speed S_2
+
+x3 = (3/4);
+S_3 = S_o - 300*x3 ; // Speed at 3/4 load
+Rot_losses_S_3 = (S_3/S)*Rot_losses ; // Rotational losses in W at speed S_3
+
+x4 = (5/4);
+S_4 = S_o - 300*x4 ; // Speed at 5/4 load
+Rot_losses_S_4 = (S_4/S)*Rot_losses ; // Rotational losses in W at speed S_4
+
+// case b
+I_sh = V / R_sh ; // Full-load shunt field current in A
+Ia = I_L - I_sh ; // Full-load armature current in A
+FL_variable_loss = (Ia^2)*(R_a + R_s); // Full-load variable electric losses in W
+
+x1_variable_loss = FL_variable_loss * (x1)^2 ; // Variable losses at 1/4 load
+x2_variable_loss = FL_variable_loss * (x2)^2 ; // Variable losses at 1/2 load
+x3_variable_loss = FL_variable_loss * (x3)^2 ; // Variable losses at 3/4 load
+x4_variable_loss = FL_variable_loss * (x4)^2 ; // Variable losses at 5/4 load
+
+// case c
+// Efficiency of motor = (Input - losses)/Input
+// where Input = volts*amperes*load_fraction
+// Losses = field loss + rotational losses + variable electric losses
+// Input
+Input_FL = V * I_L ; // Input in W at full load
+Input_x1 = V * I_L * x1 ; // Input in W at 1/4 load
+Input_x2 = V * I_L * x2 ; // Input in W at 1/2 load
+Input_x3 = V * I_L * x3 ; // Input in W at 3/4 load
+Input_x4 = V * I_L * x4 ; // Input in W at 5/4 load
+
+Field_loss = V * I_sh // Field loss for each of the conditions of load
+
+// Rotational losses are calculated in part a while variable electric losses in part b
+
+// Total losses
+Losses_FL = Field_loss + Rot_losses + FL_variable_loss ; // Total losses for full load
+Losses_1 = Field_loss + Rot_losses_S_1 + x1_variable_loss ; // Total losses for 1/4 load
+Losses_2 = Field_loss + Rot_losses_S_2 + x2_variable_loss ; // Total losses for 1/2 load
+Losses_3 = Field_loss + Rot_losses_S_3 + x3_variable_loss ; // Total losses for 3/4 load
+Losses_4 = Field_loss + Rot_losses_S_4 + x4_variable_loss ; // Total losses for 5/4 load
+
+// Efficiency
+eta_FL = ( (Input_FL - Losses_FL) / Input_FL ) ; // Efficiency for 1/4 load
+eta_1 = ( (Input_x1 - Losses_1) / Input_x1 ) ; // Efficiency for 1/4 load
+eta_2 = ( (Input_x2 - Losses_2) / Input_x2 ) ; // Efficiency for 1/2 load
+eta_3 = ( (Input_x3 - Losses_3) / Input_x3 ) ; // Efficiency for 3/4 load
+eta_4 = ( (Input_x4 - Losses_4) / Input_x4 ) ; // Efficiency for 5/4 load
+
+// Display the results
+disp("Example 12-4 Solution : ");
+
+printf(" \n a: Rotational loss = %d W at %d rpm(rated load)\n",Rot_losses,S);
+printf(" \n Speed at %.2f load = %d rpm ",x1 , S_1 );
+printf(" \n Rotational loss at %d rpm = %d W \n ", S_1 , Rot_losses_S_1 );
+
+printf(" \n Speed at %.2f load = %d rpm ",x2 , S_2 );
+printf(" \n Rotational loss at %d rpm = %d W \n ", S_2 , Rot_losses_S_2 );
+
+printf(" \n Speed at %.2f load = %d rpm ",x3 , S_3 );
+printf(" \n Rotational loss at %d rpm = %d W \n ", S_3 , Rot_losses_S_3 );
+
+printf(" \n Speed at %.2f load = %d rpm ",x4 , S_4 );
+printf(" \n Rotational loss at %d rpm = %d W \n ", S_4 , Rot_losses_S_4 );
+
+printf(" \n b: Full-load variable loss = %d W\n ",FL_variable_loss );
+printf(" \n Variable losses ,");
+printf(" \n at %.2f load = %.2f W ",x1 , x1_variable_loss );
+printf(" \n at %.2f load = %.2f W ",x2 , x2_variable_loss );
+printf(" \n at %.2f load = %.2f W ",x3 , x3_variable_loss );
+printf(" \n at %.2f load = %.2f W \n ",x4 , x4_variable_loss );
+
+printf(" \n c: Efficiency of motor = (Input - losses)/Input ");
+printf(" \n where\n Input = volts*amperes*load_fraction ");
+printf(" \n Losses = field loss + rotational losses + variable electric losses");
+printf(" \n Input,\n at %.2f load = %d W ",x1 , Input_x1 );
+printf(" \n at %.2f load = %d W ",x2 , Input_x2 );
+printf(" \n at %.2f load = %d W ",x3 , Input_x3 );
+printf(" \n at full load = %d W " , Input_FL );
+printf(" \n at %.2f load = %d W \n ",x4 , Input_x4 );
+
+printf(" \n Field loss for each of the conditions of load = %d W \n",Field_loss);
+printf(" \n Rotational losses are calculated in part a while variable ");
+printf(" \n electric losses in part b \n");
+
+printf(" \n Efficiency at %.2f load = %f = %.1f percent ",x1,eta_1,eta_1*100);
+printf(" \n Efficiency at %.2f load = %f = %.1f percent ",x2,eta_2,eta_2*100);
+printf(" \n Efficiency at %.2f load = %f = %.1f percent ",x3,eta_3,eta_3*100);
+printf(" \n Efficiency at full load = %f = %.1f percent ",eta_FL,eta_FL*100);
+printf(" \n Efficiency at %.2f load = %f = %.1f percent \n",x4,eta_4,eta_4*100);
+
+printf(" \n d: ________________________________________________________________________________________________________");
+printf(" \n Item \t\t\t At 1/4 load \t At 1/2 load \t At 3/4 load \t At Full-load\t At 5/4 load ");
+printf(" \n ________________________________________________________________________________________________________");
+printf(" \n Input(watts)\t\t %d \t\t %d \t\t %d \t\t %d \t %d ",Input_x1,Input_x2,Input_x3,Input_FL,Input_x4);
+printf(" \n\n Field loss(watts)\t\t %d \t\t %d \t\t %d \t\t %d \t\t %d ",Field_loss,Field_loss,Field_loss,Field_loss,Field_loss);
+printf(" \n\n Rotational losses");
+printf(" \n from part(a)(watts)\t\t %d \t\t %d \t\t %d \t\t %d \t\t %d ",Rot_losses_S_1,Rot_losses_S_2,Rot_losses_S_3,Rot_losses,Rot_losses_S_4);
+printf(" \n\n Variable electric losses");
+printf(" \n from part(b)(watts)\t\t %.2f \t %.2f \t %.2f \t %.2f \t %.2f ",x1_variable_loss,x2_variable_loss,x3_variable_loss,FL_variable_loss,x4_variable_loss);
+printf(" \n\n Total losses(watts)\t\t %.2f \t %.2f \t %.2f \t %.2f \t %.2f ",Losses_1,Losses_2,Losses_3,Losses_FL,Losses_4);
+printf(" \n ________________________________________________________________________________________________________");
+printf(" \n Efficiency η(percent)\t %.1f \t\t %.1f \t\t %.1f \t\t %.1f \t\t %.1f ",eta_1*100,eta_2*100,eta_3*100,eta_FL*100,eta_4*100);
+printf(" \n ________________________________________________________________________________________________________");
diff --git a/1092/CH12/EX12.5/Example12_5.sce b/1092/CH12/EX12.5/Example12_5.sce new file mode 100755 index 000000000..efad7063e --- /dev/null +++ b/1092/CH12/EX12.5/Example12_5.sce @@ -0,0 +1,47 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 10000 ; // Power rating of the shunt generator in W
+V = 230 ;// Voltage rating of the shunt generator in volt
+S = 1750 ; // Speed in rpm of the shunt generator
+R_a = 0.2 ; // Armature resistance
+// Calculated values from Ex.12-1
+P_r = 489.2 ; // Shunt generator rotational losses in W
+Vf_If = 230 ; // Shunt field circuit loss in W
+I_a_rated = 44.5 ; // Rated armature current in A
+
+// Calculations
+// case a
+I_a = sqrt( (Vf_If + P_r) / R_a ); // Armature current in A for max.efficiency
+
+// case b
+LF = I_a / I_a_rated ; // Load fraction
+LF_percent = LF*100 ; // Load fraction in percent
+
+// case c
+P_k = Vf_If + P_r ;
+eta_max = (P*LF)/( (P*LF) + (Vf_If + P_r) + P_k ) * 100; // Maximum efficiency
+
+// case d
+// subscript d for LF indicates case d
+LF_d = sqrt(P_k/(I_a_rated^2*R_a)) ; // Load fraction from fixed losses and rated variable losses
+
+// Display the results
+disp("Example 12-5 Solution : ");
+
+printf(" \n a: Armature current for max.efficiency :\n I_a = %.f A \n",I_a);
+
+printf(" \n b: Load fraction :\n L.F. = %.1f percent = %.3f*rated \n",LF_percent,LF);
+
+printf(" \n c: Maximum efiiciency :\n η = %.2f percent \n",eta_max);
+
+printf(" \n d: Load fraction from fixed losses and rated variable losses :");
+printf(" \n L.F. = %.3f*rated",LF_d);
diff --git a/1092/CH12/EX12.6/Example12_6.sce b/1092/CH12/EX12.6/Example12_6.sce new file mode 100755 index 000000000..7a604ecf4 --- /dev/null +++ b/1092/CH12/EX12.6/Example12_6.sce @@ -0,0 +1,52 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 240 ; // Voltage rating of dc shunt motor in volt
+P_hp = 5 ; // Power rating of dc shunt motor in hp
+S = 1100 ; // Speed in rpm of the dc shunt motor
+R_a = 0.4 ; // Armture resistance in ohm
+R_f = 240 ; // Field resistance in ohm
+I_L = 20 ; // Rated line current in A
+
+// Calculations
+// Preliminary calculations
+V_f = V ; // Voltage across field winding in volt
+I_f = V_f / R_f ; // Field current in A
+I_a = I_L - I_f ; // Armature current in A
+P_o = P_hp * 746 ; // Power rating of dc shunt motor in W
+V_a = V ; // Voltage across armature in volt
+E_c_fl = V_a - I_a*R_a ; // back EMF in volt
+
+// case a
+E_c = E_c_fl ;
+P_d = E_c * I_a ; // Power developed by the armature in W
+
+// case b
+P_r = P_d - P_o ; // Full-load rotational losses in W
+
+// case c
+P_in = V*I_L ; // Input power in W
+eta = (P_o/P_in)*100 ; // Full-load efficiency
+
+// Display the results
+disp("Example 12-6 Solution : ");
+
+printf(" \n Preliminary calculations using nameplate data : ");
+printf(" \n Field current : I_f = %d A \n ",I_f);
+printf(" \n Armature current : I_a = %d A \n ",I_a);
+printf(" \n P_o = %d W ",P_o );
+printf(" \n E_c(fl) = %.1f V \n",E_c_fl);
+
+printf(" \n a: Power developed by the armature :\n P_d = %.1f W \n",P_d);
+
+printf(" \n b: Full-load rotational losses :\n P_r = %.1f W \n",P_r);
+
+printf(" \n c: Full-laod efficiency :\n η = %.1f percent ",eta );
diff --git a/1092/CH12/EX12.7/Example12_7.sce b/1092/CH12/EX12.7/Example12_7.sce new file mode 100755 index 000000000..f574bfab9 --- /dev/null +++ b/1092/CH12/EX12.7/Example12_7.sce @@ -0,0 +1,73 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 240 ; // Voltage rating of dc shunt motor in volt
+P_hp = 25 ; // Power rating of dc shunt motor in hp
+S = 1100 ; // Speed in rpm of the dc shunt motor
+R_a = 0.15 ; // Armture resistance in ohm
+R_f = 80 ; // Field resistance in ohm
+I_L = 89 ; // Rated line current in A
+
+// Calculations
+// Preliminary calculations
+V_f = V ; // Voltage across field winding in volt
+I_f = V_f / R_f ; // Field current in A
+I_a = I_L - I_f ; // Armature current in A
+P_o = P_hp * 746 ; // Power rating of dc shunt motor in W
+V_a = V ; // Voltage across armature in volt
+E_c_fl = V_a - I_a*R_a ; // back EMF in volt
+
+// case a
+E_c = E_c_fl ;
+P_d = E_c * I_a ; // Power developed by the armature in W
+
+// case b
+P_r = P_d - P_o ; // Full-load rotational losses in W
+
+// case c
+P_in = V*I_L ; // Input power in W
+eta_fl = (P_o/P_in)*100 ; // Full-load efficiency
+
+// case d
+P_k = V_f*I_f + P_r ; // Total constant losses in W
+
+// case e
+Ia = sqrt(P_k/R_a); // Armature current in A from maximum efficiency
+
+// case f
+LF = Ia / I_a ; // Load fraction at which max.efficiency is produced
+
+// case g
+rated_input = V*I_L ;
+eta_max = ( (LF*rated_input) - 2*P_k ) / (LF*rated_input) * 100; // Maximum efficiency
+
+// Display the results
+disp("Example 12-7 Solution : ");
+
+printf(" \n Field current : I_f = %d A \n ",I_f);
+printf(" \n Armature current : I_a = %d A \n ",I_a);
+printf(" \n P_o = %d W \n",P_o );
+printf(" \n E_c(fl) = %.1f V \n",E_c_fl);
+
+printf(" \n a: Power developed by the armature :\n P_d = %.1f W \n",P_d);
+
+printf(" \n b: Full-load rotational losses :\n P_r = %.1f W \n",P_r);
+
+printf(" \n c: Full-laod efficiency :\n η = %.1f percent \n ",eta_fl );
+
+printf(" \n d: Total constant losses :\n P_k = %.1f W \n",P_k);
+
+printf(" \n e: Armature current from maximum efficiency :\n I_a = %.1f A\n ",Ia);
+
+printf(" \n f: L.F. = %.1f \n ",LF);
+
+printf(" \n g: η_max = %.1f percent",eta_max);
+
diff --git a/1092/CH12/EX12.8/Example12_8.sce b/1092/CH12/EX12.8/Example12_8.sce new file mode 100755 index 000000000..26d2bafb1 --- /dev/null +++ b/1092/CH12/EX12.8/Example12_8.sce @@ -0,0 +1,71 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 240 ; // Voltage rating of dc shunt motor in volt
+P_hp = 5 ; // Power rating of dc shunt motor in hp
+S_fl = 1100 ; // Speed in rpm of the dc shunt motor
+R_a = 0.4 ; // Armture resistance in ohm
+R_f = 240 ; // Field resistance in ohm
+eta = 0.75 ; // Full-load efficiency
+
+// Calculations
+// case a
+V_L = V ; // Load voltage
+P_o = P_hp * 746 ; // Power rating of dc shunt motor in W
+I_L = P_o / (eta*V_L); // Rated input line current in A
+
+V_f = V ; // Voltage across field winding in volt
+I_f = V_f / R_f ; // Field current in A
+I_a = I_L - I_f ; // Rated armature current in A
+
+// case b
+V_a = V ; // Voltage across armature in volt
+E_c = V_a - I_a*R_a ; // back EMF in volt
+P_d = E_c * I_a ; // Power developed by the armature in W
+
+// case c
+P_r = P_d - P_o ; // Rotational losses in W at rated load
+
+// case d
+// At no-load
+P_o_nl = 0 ;
+P_r_nl = P_r ; // Rotational losses in W at no load
+P_d_nl = P_r_nl ;
+
+// case e
+I_a_nl = P_d_nl / V_a ; // No-load armature current in A
+
+// case f
+E_c_nl = V ; // No-load voltage in volt
+E_c_fl = E_c ; // Full-load voltage in volt
+S_nl = (E_c_nl / E_c_fl)*S_fl ; // No-load speed in rpm
+
+// case g
+SR = (S_nl - S_fl)/S_fl * 100 ; // Speed regulation
+
+// Display the results
+disp("Example 12-8 Solution : ");
+
+printf(" \n a: Rated input line current :\n I_L = %.2f A \n ",I_L);
+printf(" \n Rated armature current :\n I_a = %.2f A \n ",I_a );
+
+printf(" \n b: E_c = %.1f V \n ",E_c );
+printf(" \n Power developed by the armature at rated load :\n P_d = %d W \n ",P_d);
+
+printf(" \n c: Rotational losses at rated load :\n P_r = %d W \n ", P_r );
+
+printf(" \n d: At no-load, P_o = %d W ; therefore\n\t\tP_d = P_r = %d W \n",P_o_nl,P_r);
+
+printf(" \n e: No-load armature current :\n I_a(nl) = %.2f A \n ",I_a_nl );
+
+printf(" \n f: No-load speed :\n S_nl = %.f rpm \n ",S_nl );
+
+printf(" \n g: Speed regulation :\n SR = %.1f percent ",SR );
diff --git a/1092/CH12/EX12.9/Example12_9.sce b/1092/CH12/EX12.9/Example12_9.sce new file mode 100755 index 000000000..26f891b56 --- /dev/null +++ b/1092/CH12/EX12.9/Example12_9.sce @@ -0,0 +1,79 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
+// Example 12-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 240 ; // Voltage rating of dc shunt motor in volt
+I_L = 55 ; // Rated line current in A
+S = 1200 ; // Speed in rpm of the dc shunt motor
+P_r = 406.4 ; // Rotational losses in W at rated load
+R_f = 120 ; // Field resistance in ohm
+R_a = 0.4 ; // Armture resistance in ohm
+
+// Calculations
+// case a
+
+V_f = V ; // Voltage across field winding in volt
+I_f = V_f / R_f ; // Field current in A
+I_a = I_L - I_f ; // Rated armature current in A
+
+V_a = V ; // Voltage across armature in volt
+E_c = V_a - I_a*R_a ; // back EMF in volt
+P_d = E_c * I_a ; // Power developed by the armature in W
+
+// case b
+P_o = P_d - P_r ; // Rated output power in W
+P_o_hp = P_o / 746 ; // Rated output power in hp
+
+// case c
+T_o = (P_o_hp * 5252)/S ; // C in lb-ft
+T_o_Nm = T_o * (1.356); // Rated output torque in N-m
+
+// case d
+P_in = V*I_L ; // Input power in W
+eta = (P_o/P_in)*100 ; // Efficiency at rated load
+
+// case e
+// At no-load
+P_o_nl = 0 ;
+P_r_nl = P_r ; // Rotational losses in W at no load
+P_d_nl = P_r_nl ;
+
+I_a_nl = P_d_nl / V_a ; // No-load armature current in A
+
+E_c_nl = V ; // No-load voltage in volt
+E_c_fl = E_c ; // Full-load voltage in volt
+S_fl = S ; // Full-load speed in rpm
+S_nl = (E_c_nl / E_c_fl)*S_fl ; // No-load speed in rpm
+
+// case f
+SR = (S_nl - S_fl)/S_fl * 100 ; // Speed regulation
+
+// Display the results
+disp("Example 12-9 Solution : ");
+
+printf(" \n a: E_c = %.1f V \n ",E_c );
+printf(" \n Power developed by the armature at rated load :\n P_d = %.1f W \n ",P_d);
+
+printf(" \n b: Rated output power :\n P_o = %d W \n ", P_o );
+printf(" \n P_o = %d hp \n ",P_o_hp);
+
+printf(" \n c: Rated output torque :\n T_o = %.2f lb-ft ",T_o);
+printf(" \n T_o = %.f N-m \n ",T_o_Nm );
+
+printf(" \n d: Efficiency at rated load :\n η = %.1f percent \n ",eta );
+
+printf(" \n e: At no-load, P_o = %d W ; therefore\n\t\tP_d = P_r = EcIa ≅ VaIa = %.1f W \n",P_o_nl,P_r);
+printf(" \n No-load armature current :\n I_a(nl) = %.3f A \n ",I_a_nl );
+printf(" \n No-load speed :\n S_nl = %f ≃ %.f rpm \n ",S_nl,S_nl );
+
+printf(" \n f: Speed regulation :\n SR = %.1f percent ",SR );
+
+printf(" \n Variation in SR is due to non-approximation of S_nl = %f rpm",S_nl);
+printf(" \n while calculating SR in scilab .")
diff --git a/1092/CH13/EX13.1/Example13_1.sce b/1092/CH13/EX13.1/Example13_1.sce new file mode 100755 index 000000000..751bee84a --- /dev/null +++ b/1092/CH13/EX13.1/Example13_1.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// MOTOR(class A insulation ) is operated for 6 hrs
+T = 125 ; // Temperature in degree celsius recorded by the embedded detectors
+life_orig = 10 ; // Life in years of the motor (standard)
+
+// Calculations
+delta_T = T - 105 ; // Positive temperature difference between the given
+// max hottest spot temperature of its insulation and the ambient temperature recorded.
+// 105 is chosen from table 13-1(class A insulation)
+R = 2 ^ (delta_T/10); // Life reduction factor
+
+Life_calc = life_orig / R ; // Reduced life expectancy of the motor in years
+
+// Display the results
+disp("Example 13-1 Solution : ");
+printf(" \n Life reduction factor : R = %d \n ",R );
+printf(" \n Reduced life expectancy of the motor : Life_calc = %.1f years",Life_calc);
diff --git a/1092/CH13/EX13.10/Example13_10.sce b/1092/CH13/EX13.10/Example13_10.sce new file mode 100755 index 000000000..ff3b925cb --- /dev/null +++ b/1092/CH13/EX13.10/Example13_10.sce @@ -0,0 +1,31 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// single phase alternator
+V_orig = 500 ; // Rated voltage of the alternator in volt
+kVA_orig = 20 ; // Rated power of the alternator in kVA
+I = 40 ; // Rated current of the alternator in A
+R = 2 ; // Armature resistance in ohm
+X = 15 ; // Armature reactance in ohm
+
+V_new = 5000 ; // New voltage of the alternator in volt
+kVA_new = 100 ; // New power of the alternator in kVA
+
+// Calculated armature impedance from Ex.13-9c
+Z_pu_orig = 1.211 ; // original per-unit value of armature impedance in p.u
+
+// Calculation
+Z_pu_new = Z_pu_orig * (kVA_new/kVA_orig) * (V_orig/V_new)^2 ;
+// new per-unit value of armature impedance in p.u
+
+// Display the results
+disp("Example 13-10 Solution : ");
+printf(" \n New per-unit value of armature impedance\n Z_pu(new) = %.5f p.u",Z_pu_new);
diff --git a/1092/CH13/EX13.11/Example13_11.sce b/1092/CH13/EX13.11/Example13_11.sce new file mode 100755 index 000000000..77b115ae2 --- /dev/null +++ b/1092/CH13/EX13.11/Example13_11.sce @@ -0,0 +1,30 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase distribution system
+V = 2300 ; // Line voltage of 3-phase distribution system in volt
+V_p = 1328 ; // Phase voltage of 3-phase distribution system in volt
+
+V_b = 69000 ; // Common base line voltage in volt
+V_pb = 39840 ; // Common base phase voltage in volt
+
+// Calculations
+// case a
+V_pu_line = V / V_b ; // Distribution system p.u line voltage
+
+// case a
+V_pu_phase = V_p / V_pb ; // Distribution system p.u phase voltage
+
+// Display the results
+disp("Example 13-11 Solution : ");
+printf(" \n a: Distribution system p.u line voltage :\n V_pu = %.2f p.u\n",V_pu_line);
+
+printf(" \n b: Distribution system p.u phase voltage :\n V_pu = %.2f p.u\n",V_pu_phase);
diff --git a/1092/CH13/EX13.12/Example13_12.sce b/1092/CH13/EX13.12/Example13_12.sce new file mode 100755 index 000000000..9441448c1 --- /dev/null +++ b/1092/CH13/EX13.12/Example13_12.sce @@ -0,0 +1,70 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+VA_b = 50 ; // Base power rating of the 3-phase Y-connected alternator in MVA
+V_b = 25 ; // Base voltage of the 3-phase Y-connected alternator in kV
+X_pu = 1.3 ; // per unit value of synchronous reactance
+R_pu = 0.05 ; // per unit value of resistance
+
+// Calculations
+// case a
+// subscript 1 for Z_b indicates method 1 for finding Z_b
+Z_b1 = (V_b)^2 / VA_b ; // Base impedance in ohm
+
+// subscript 2 for Z_b indicates method 2 for finding Z_b
+S_b = VA_b ; // Base power rating of the 3-phase Y-connected alternator in MVA
+I_b = (S_b)/V_b ; // Base current in kA
+Z_b2 = V_b / I_b ; // Base impedance in ohm
+
+// case b
+Z_b = Z_b1; // Base impedance in ohm
+X_s = X_pu * Z_b ; // Actual value of synchronous reactance per phase in ohm
+
+// case c
+R_a = R_pu * Z_b ; // Actual value of armature stator resistance per phase in ohm
+
+// case d
+// subscript 1 for Z_s indicates method 1 for finding Z_s
+Z_s1 = R_a + %i*X_s ; // Synchronous impedance per phase in ohm
+Z_s1_m = abs(Z_s1);//Z_s1_m = magnitude of Z_s1 in ohm
+Z_s1_a = atan(imag(Z_s1) /real(Z_s1))*180/%pi;//Z_s1_a=phase angle of Z_s1 in degrees
+
+// subscript 2 for Z_s indicates method 2 for finding Z_s
+Z_pu = R_pu + %i*X_pu ; // per unit value of impedance
+Z_s2 = Z_pu * Z_b ; // Synchronous impedance per phase in ohm
+Z_s2_m = abs(Z_s2);//Z_s2_m = magnitude of Z_s2 in ohm
+Z_s2_a = atan(imag(Z_s2) /real(Z_s2))*180/%pi;//Z_s2_a=phase angle of Z_s2 in degrees
+
+// case e
+S = S_b ; // Base power rating of the 3-phase Y-connected alternator in MVA
+P = S * R_pu ; // Full-load copper losses for all three phases in MW
+
+// Display the results
+disp("Example 13-12 Solution : ");
+
+printf(" \n a: Base impedance(method 1): \n Z_b = %.1f ohm\n",Z_b1);
+printf(" \n Base impedance(method 2) : ");
+printf(" \n I_b = %d kA \n Z_b = %.1f ohm\n",I_b,Z_b2);
+
+printf(" \n b: Actual value of synchronous reactance per phase : ");
+printf(" \n X_s in ohm = ");disp(%i*X_s);
+
+printf(" \n c: Actual value of armature stator resistance per phase : ");
+printf(" \n R_a = %.3f ohm \n ",R_a );
+
+printf(" \n d: Synchronous impedance per phase (method 1): ");
+printf(" \n Z_s in ohm = ");disp(Z_s1);
+printf(" \n Z_s = %.2f <%.1f ohm\n",Z_s1_m,Z_s1_a);
+printf(" \n Synchronous impedance per phase (method 2) : ");
+printf(" \n Z_s in ohm = ");disp(Z_s2);
+printf(" \n Z_s = %.2f <%.1f ohm\n",Z_s2_m,Z_s2_a);
+
+printf(" \n e: Full-load copper losses for all 3 phases : \n P = %.1f MW",P);
diff --git a/1092/CH13/EX13.2/Example13_2.sce b/1092/CH13/EX13.2/Example13_2.sce new file mode 100755 index 000000000..3f962ee53 --- /dev/null +++ b/1092/CH13/EX13.2/Example13_2.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// MOTOR(class A insulation ) is operated for 6 hrs
+T = 75 ; // Temperature in degree celsius recorded by the embedded detectors
+life_orig = 10 ; // Life in years of the motor (standard)
+
+// Calculations
+delta_T = 105 - T ; // Positive temperature difference between the given
+// max hottest spot temperature of its insulation and the ambient temperature recorded.
+// 105 is chosen from table 13-1 (class A insulation)
+E = 2 ^ (delta_T/10); // Life extension factor
+
+Life_calc = life_orig * E ; // Increased life expectancy of the motor in years
+
+// Display the results
+disp("Example 13-2 Solution : ");
+printf(" \n Life extension factor : E = %d \n ",E );
+printf(" \n Increased life expectancy of the motor : Life_calc = %d years ",Life_calc);
diff --git a/1092/CH13/EX13.3/Example13_3.sce b/1092/CH13/EX13.3/Example13_3.sce new file mode 100755 index 000000000..4cbc66eb6 --- /dev/null +++ b/1092/CH13/EX13.3/Example13_3.sce @@ -0,0 +1,29 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Class A insulation
+T_A = 105 ; // Temperature in degree celsius recorded by the embedded detectors
+life_orig = 5 ; // Life in years of the motor (standard)
+// Class B insulation
+T_B = 130 ; // Temperature in degree celsius recorded by the embedded detectors
+
+// Calculations
+delta_T = T_B - T_A ; // Positive temperature difference betw the given
+// max hottest spot temperature of its insulation and the ambient temperature recorded.
+// T_A and T_B are chosen from table 13-1
+E = 2 ^ (delta_T/10); // Life extension factor
+
+Life_calc = life_orig * E ; // Increased life expectancy of the motor in years
+
+// Display the results
+disp("Example 13-3 Solution : ");
+printf(" \n Life extension factor : E = %.2f \n ",E );
+printf(" \n Increased life expectancy of the motor : Life_calc = %.1f years ",Life_calc);
diff --git a/1092/CH13/EX13.4/Example13_4.sce b/1092/CH13/EX13.4/Example13_4.sce new file mode 100755 index 000000000..8f3db574a --- /dev/null +++ b/1092/CH13/EX13.4/Example13_4.sce @@ -0,0 +1,45 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 25 ; // Rated power of SCIM in hp
+// class B insulation
+T_ambient = 40 ; // Standard ambient temperature recorded by the embedded hot-spot detectors in degree celsius
+T_hottest = 115 ; // Hottest-spot winding temperature recorded by the embedded hot-spot detectors in degree celsius
+
+// Calculations
+// case a
+// from table 13-1 allowable temperature rise in 90 degree celsius
+
+// case b
+T_rise = T_hottest - T_ambient ; // Actual temperature rise for the insulation type used in degree celsius
+
+// case c
+P_f = P_o * (90/T_rise); // Approximate power to the motor that can be delivered at T_rise
+
+// case d
+// same as P_f
+
+// case e
+// answer from case a
+
+// Display the results
+disp("Example 13-4 Solution : ");
+printf(" \n a: The allowable temperature rise for the ");
+printf(" \n insulation type used = 90 degree celsius(from table 13-1)\n");
+
+printf(" \n b: The actual temperature rise for the insulation type used = %d degree celsius\n",T_rise);
+
+printf(" \n c: The approximate power to the motor that can be delivered at T_rise");
+printf(" \n P_f = %d hp\n",P_f);
+
+printf(" \n d: Power rating that may be stamped on the nameplate = %d hp(subject to test at this load) \n ",P_f);
+
+printf(" \n e: The temperature rise that must be stamped on the nameplate = 90 degree celsius");
diff --git a/1092/CH13/EX13.5/Example13_5.sce b/1092/CH13/EX13.5/Example13_5.sce new file mode 100755 index 000000000..913e7ed13 --- /dev/null +++ b/1092/CH13/EX13.5/Example13_5.sce @@ -0,0 +1,45 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 50 ; // Power rating of the WRIM in hp
+// Class F insulation
+T_hottest = 160 ; // Hottest-spot winding temperature recorded by the embedded
+// hot-spot detectors in degree celsius
+T_ambient = 40 ; // Standard ambient temperature recorded by the embedded
+// hot-spot detectors in degree celsius
+P_f_a = 40 ; // Power rating of load a in hp
+P_f_b = 55 ; // Power rating of load a in hp
+
+// Calculations
+// case a
+delta_T_o = T_hottest - T_ambient ; // Temperature rise for the insulation type
+// used in degree celsius
+
+// subscript a in delta_T_f ,P_f_a and T_f indicates case a
+delta_T_f_a = (P_f_a/P_o)*delta_T_o ; // final temperature rise in degree celsius
+T_f_a = delta_T_f_a + T_ambient ; // Approximate final hot-spot temperature in degree celsius
+
+// case b
+// subscript b in delta_T_f ,P_f and T_f indicates case b
+delta_T_f_b = (P_f_b/P_o)*delta_T_o ; // final temperature rise in degree celsius
+T_f_b = delta_T_f_b + T_ambient ; // Approximate final hot-spot temperature in degree celsius
+
+// Display the results
+disp("Example 13-5 Solution : ");
+printf(" \n a: ΔT_o = %d degree celsius ",delta_T_o);
+printf(" \n ΔT_f = %d degree celsius ",delta_T_f_a);
+printf(" \n T_f = %d degree celsius \n",T_f_a);
+
+printf(" \n b: ΔT_f = %d degree celsius ",delta_T_f_b);
+printf(" \n T_f = %d degree celsius \n",T_f_b);
+printf(" \n Yes,motor life is reduced at the 110 percent motor load because");
+printf(" \n the allowable maximum hot-spot motor temperature for Class F");
+printf(" \n insulation is 155 degree celsius.");
diff --git a/1092/CH13/EX13.6/Example13_6.sce b/1092/CH13/EX13.6/Example13_6.sce new file mode 100755 index 000000000..75903fe46 --- /dev/null +++ b/1092/CH13/EX13.6/Example13_6.sce @@ -0,0 +1,33 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 55 ; // Power rating of the WRIM in hp
+T_ambient = 40 ; // Standard ambient temperature recorded by the embedded
+// hot-spot detectors in degree celsius
+life_orig = 10 ; // Life in years of the motor (standard)
+
+// Calculated data from Ex.13-5b
+T_f = 172 ; // Approximate final hot-spot temperature in degree celsius
+
+// Calculations
+delta_T = T_f - 155 ; // Positive temperature difference betw the given
+// max hottest spot temperature of its insulation and the ambient temperature recorded.
+// 155 is chosen from table 13-1(class F insulation)
+
+R = 2 ^ (delta_T/10); // Life reduction factor
+
+Life_calc = life_orig / R ; // Reduced life expectancy of the motor in years
+
+// Display the results
+disp("Example 13-6 Solution : ");
+printf(" \n From Ex.13-5b,T_f = %d degree celsius\n",T_f);
+printf(" \n Life reduction factor : R = %.2f \n ",R );
+printf(" \n Reduced life expectancy of the motor : Life_calc = %.2f years",Life_calc);
diff --git a/1092/CH13/EX13.7/Example13_7.sce b/1092/CH13/EX13.7/Example13_7.sce new file mode 100755 index 000000000..710731bf6 --- /dev/null +++ b/1092/CH13/EX13.7/Example13_7.sce @@ -0,0 +1,29 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 200 ; // Power rating of the test motor in hp
+t1 = 5 ; // time duration in minutes for which test motor is operated at 200 hp
+t2 = 5 ; // time duration in minutes for which test motor is operated at 20 hp
+t3 = 10 ; // time duration in minutes for which test motor is operated at 100 hp
+
+// Calculation
+rms_hp = sqrt( ( (200^2)*t1 + (20^2)*t2 + (100^2)*t3 )/(t1 + t2 + t3 + 10/3) );
+// Horsepower required for intermittent varying load
+
+// Display the results
+disp("Example 13-7 Solution : ");
+printf(" \n Horsepower required for intermittent varying load is : ");
+printf(" \n rms hp = %.f hp \n ",rms_hp);
+
+printf(" \n A 125 hp motor would be selected because that is the nearest larger");
+printf(" \n commercial standard rating.This means that the motor would operate ");
+printf(" \n with a 160 percent overload (at 200 hp) for 5 minutes,or 1/6th of ")
+printf(" \n its total duty cycle.");
diff --git a/1092/CH13/EX13.8/Example13_8.sce b/1092/CH13/EX13.8/Example13_8.sce new file mode 100755 index 000000000..d754c00fc --- /dev/null +++ b/1092/CH13/EX13.8/Example13_8.sce @@ -0,0 +1,38 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 120 ; // Rated output voltage in volt of separately excited dc generator
+I = 100 ; // Rated output current in A of separately excited dc generator
+R = 0.1 ; // Armature resistance in ohm
+
+// Calculations
+// case a
+V_b = V ; // base voltage in volt
+
+// case b
+I_b = I ; // base current in A
+
+// case c
+R_b = V_b / I_b ; // base resistance in ohm
+
+// case d
+R_pu = R / R_b ; // per-unit value of armature resistance in p.u
+
+// Display the results
+disp("Example 13-8 Solution : ");
+
+printf(" \n a: Base voltage \n V_b = %d V \n ", V_b );
+
+printf(" \n b: Base current \n I_b = %d A \n ", I_b );
+
+printf(" \n c: Base resistance \n R_b = %.1f ohm \n ", R_b );
+
+printf(" \n d: Per-unit value of armature resistance\n R_p.u = %.3f p.u \n ", R_pu );
diff --git a/1092/CH13/EX13.9/Example13_9.sce b/1092/CH13/EX13.9/Example13_9.sce new file mode 100755 index 000000000..3643c9a36 --- /dev/null +++ b/1092/CH13/EX13.9/Example13_9.sce @@ -0,0 +1,51 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY
+// Example 13-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// single phase alternator
+V = 500 ; // Rated voltage of the alternator in volt
+P = 20 ; // Rated power of the alternator in kVA
+I = 40 ; // Rated current of the alternator in A
+R = 2 ; // Armature resistance in ohm
+X = 15 ; // Armature reactance in ohm
+
+// Calculations
+// case a
+V_b = V ; // base voltage in volt
+I_b = I ; // base current in A
+R_pu = (R*I_b)/V_b ; // per-unit value of armature resistance in p.u
+
+// case b
+jX_pu = (X*I_b)/V_b ; // per-unit value of armature reactance in p.u
+
+// case c
+// subscript 1 indicates method 1 for finding Z_p.u
+Z_pu1 = R_pu + %i*(jX_pu); // per-unit value of armature impedance in p.u
+Z_pu1_m = abs(Z_pu1);//Z_pu1_m = magnitude of Z_pu1 in p.u
+Z_pu1_a = atan(imag(Z_pu1) /real(Z_pu1))*180/%pi;//Z_pu1_a=phase angle of Z_pu1 in degrees
+
+// subscript 2 indicates method 2 for finding Z_p.u
+Z_pu2 = (R + %i*X)*(I/V); // per-unit value of armature impedance in p.u
+Z_pu2_m = abs(Z_pu2);//Z_pu2_m = magnitude of Z_pu2 in p.u
+Z_pu2_a = atan(imag(Z_pu2) /real(Z_pu2))*180/%pi;//Z_pu2_a=phase angle of Z_pu2 in degrees
+
+// Display the results
+disp("Example 13-9 Solution : ");
+
+printf(" \n a: Armature resistance per unit value\n R_p.u = %.2f p.u \n",R_pu);
+
+printf(" \n b: Armature reactance per unit value\n jX_p.u in p.u = ");disp(%i*jX_pu);
+
+printf(" \n c: Armature impedance per unit value\n");
+printf(" \n (method 1)\n Z_p.u in p.u = ");disp(Z_pu1);
+printf(" \n Z_p.u = %.3f <%.1f p.u \n",Z_pu1_m,Z_pu1_a );
+
+printf(" \n (method 2)\n Z_p.u in p.u = ");disp(Z_pu2);
+printf(" \n Z_p.u = %.3f <%.1f p.u \n",Z_pu2_m,Z_pu2_a );
diff --git a/1092/CH14/EX14.1/Example14_1.sce b/1092/CH14/EX14.1/Example14_1.sce new file mode 100755 index 000000000..921585477 --- /dev/null +++ b/1092/CH14/EX14.1/Example14_1.sce @@ -0,0 +1,37 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data for Step -down transformer
+N_1 = 500 ; // Number of turns in the primary
+N_2 = 100 ; // Number of turns in the secondary
+I_2 = 12 ; // Load (Secondary) current in A
+
+// Calculations
+// case a
+alpha = N_1 / N_2 ; // Transformation ratio
+
+// case b
+I_1 = I_2 / alpha ; // Load component of primary current in A
+
+// case c
+// sunscript c for alpha indicates case c
+// For step up transformer , using above given data
+N1 = 100 ; // Number of turns in the primary
+N2 = 500 ; // Number of turns in the secondary
+alpha_c = N1 / N2 ; // Transformation ratio
+
+// Display the results
+disp("Example 14-1 Solution : ");
+
+printf(" \n a: Transformation ratio(step-down transformer) :\n α = %d\n",alpha);
+
+printf(" \n b: Load component of primary current : \n I_1 = %.1f A \n",I_1);
+
+printf(" \n c: Transformation ratio(step-up transformer) :\n α = %.1f",alpha_c);
diff --git a/1092/CH14/EX14.10/Example14_10.sce b/1092/CH14/EX14.10/Example14_10.sce new file mode 100755 index 000000000..69f2bafee --- /dev/null +++ b/1092/CH14/EX14.10/Example14_10.sce @@ -0,0 +1,34 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data(from Example 14-9)
+kVA = 1 ; // kVA rating of the transformer
+V_1 = 220 ; // Primary voltage in volt
+V_2 = 110 ; // Secondary voltage in volt
+f_o = 400 ; // Frequency in Hz
+f_f = 60 ; // Frequency in Hz for which the transformer is to be used
+P_orig = 10 ; // Original iron losses of the transformer in W
+
+// Calculations
+// consider only ratio of frequencies for calculating B
+B = f_o / f_f ; // flux density
+
+P_iron = (P_orig)*(B^2); // Iron losses in W
+
+// Display the results
+disp("Example 14-10 Solution : ");
+
+printf(" \n Since E = k*f*B_m and the same primary voltage is applied to the ");
+printf(" \n transformer at reduced frequency, the final flux density B_mf ");
+printf(" \n increased significantly above its original maximum permissible ");
+printf(" \n value B_mo to :\n B_mf = B_mo * (f_o/f_f) = %.2fB_mo \n ",B );
+
+printf(" \n Since the iron losses vary approximately as the square of the flux density :");
+printf(" \n P_iron = %d W ",P_iron );
diff --git a/1092/CH14/EX14.11/Example14_11.sce b/1092/CH14/EX14.11/Example14_11.sce new file mode 100755 index 000000000..0e6825525 --- /dev/null +++ b/1092/CH14/EX14.11/Example14_11.sce @@ -0,0 +1,66 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 500 ; // kVA rating of the step-down transformer
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+f = 60 ; // Frequency in Hz
+r_1 = 0.1 ; // Primary winding resistance in ohm
+x_1 = 0.3 ; // Primary winding reactance in ohm
+r_2 = 0.001 ; // Secondary winding resistance in ohm
+x_2 = 0.003 ; // Secondary winding reactance in ohm
+
+// Calculations
+alpha = V_1 / V_2 ; // Transformation ratio
+// case a
+I_2 = (kVA*1000) / V_2 ; // Secondary current in A
+I_1 = I_2 / alpha ; // Primary current in A
+
+// case b
+Z_2 = r_2 + %i*(x_2); // Secondary internal impedance in ohm
+Z_2_m = abs(Z_2);//Z_2_m=magnitude of Z_2 in ohm
+Z_2_a = atan(imag(Z_2) /real(Z_2))*180/%pi;//Z_2_a=phase angle of Z_2 in degrees
+
+Z_1 = r_1 + %i*(x_1); // Primary internal impedance in ohm
+Z_1_m = abs(Z_1);//Z_1_m=magnitude of Z_1 in ohm
+Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees
+
+// case c
+I_2_Z_2 = I_2 * Z_2_m ; // Secondary internal voltage drop in volt
+I_1_Z_1 = I_1 * Z_1_m ; // Primary internal voltage drop in volt
+
+// case d
+E_2 = V_2 + I_2_Z_2 ; // Secondary induced voltage in volt
+E_1 = V_1 - I_1_Z_1 ; // Primary induced voltage in volt
+
+// case e
+ratio_E = E_1 / E_2 ; // ratio of primary to secondary induced voltage
+ratio_V = V_1 / V_2 ; // ratio of primary to secondary terminal voltage
+
+// Display the results
+disp("Example 14-11 Solution : ");
+
+printf(" \n a: Secondary current :\n I_2 = %.f A \n ",I_2 );
+printf(" \n Primary current :\n I_1 = %.1f A \n ",I_1 );
+
+printf(" \n b: Secondary internal impedance : \n Z_2 in ohm = ");disp(Z_2);
+printf(" \n Z_2 = %f <%.2f ohm \n ",Z_2_m , Z_2_a );
+printf(" \n Primary internal impedance : \n Z_1 in ohm = ");disp(Z_1);
+printf(" \n Z_1 = %f <%.2f ohm \n ",Z_1_m , Z_1_a );
+
+printf(" \n c: Secondary internal voltage drop :\n I_2*Z_2 = %.2f V \n ",I_2_Z_2);
+printf(" \n Primary internal voltage drop :\n I_1*Z_1 = %.2f V \n ",I_1_Z_1);
+
+printf(" \n d: Secondary induced voltage :\n E_2 = %.2f V \n",E_2 );
+printf(" \n Primary induced voltage :\n E_1 = %.2f V \n",E_1 );
+
+printf(" \n e: Ratio of E_1/E_2 = %.2f = α = N_1/N_2 \n",ratio_E );
+printf(" \n But V_1/V_2 = %d ",ratio_V );
diff --git a/1092/CH14/EX14.12/Example14_12.sce b/1092/CH14/EX14.12/Example14_12.sce new file mode 100755 index 000000000..b66edc909 --- /dev/null +++ b/1092/CH14/EX14.12/Example14_12.sce @@ -0,0 +1,49 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data(from Example 14-11)
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+I_2 = 2174 ; // Secondary current in A
+I_1 = 217.4 ; // Primary current in A
+// calculated values from Example 14-11
+Z_2 = 0.00316 ; // Secondary internal impedance in ohm
+Z_1 = 0.316 ; // Primary internal impedance in ohm
+
+
+// Calculations
+alpha = V_1 / V_2 ; // Transformation ratio
+// case a
+Z_L = V_2 / I_2 ; // Load impedance in ohm
+
+// case b
+Z_p = V_1 / I_1 ; // Primary input impedance in ohm
+
+Zp = (alpha)^2 * Z_L ; // Primary input impedance in ohm
+
+// Display the results
+disp("Example 14-12 Solution : ");
+
+printf(" \n a: Load impedance :\n Z_L = %.4f ohm \n ", Z_L );
+
+printf(" \n b: Primary input impedance : ");
+printf(" \n (method 1) :\n Z_p = %.2f ohm \n ",Z_p );
+printf(" \n (method 2) :\n Z_p = %.2f ohm \n ",Zp );
+
+printf(" \n c: The impedance of the load Z_L = %.4f Ω, which is much greater",Z_L);
+printf(" \n than the internal secondary impedance Z_2 = %.5f Ω .\n ",Z_2);
+printf(" \n The primary input impedance Z_p = %.2f Ω,which is much greater",Z_p);
+printf(" \n than the internal primary impedance Z_1 = %.3f Ω .\n",Z_1);
+
+printf(" \n d: It is essential for Z_L to be much greater than Z_2 so that the ");
+printf(" \n major part of the voltage produced by E_2 is dropped across the ");
+printf(" \n load impedance Z_L. As Z_L is reduced in proportion to Z_2, the ");
+printf(" \n load current increases and more voltage is dropped internally ");
+printf(" \n across Z_2.");
diff --git a/1092/CH14/EX14.13/Example14_13.sce b/1092/CH14/EX14.13/Example14_13.sce new file mode 100755 index 000000000..6825bf47a --- /dev/null +++ b/1092/CH14/EX14.13/Example14_13.sce @@ -0,0 +1,67 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-13
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 500 ; // kVA rating of the step-down transformer
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+f = 60 ; // Frequency in Hz
+r_1 = 0.1 ; // Primary winding resistance in ohm
+x_1 = 0.3 ; // Primary winding reactance in ohm
+r_2 = 0.001 ; // Secondary winding resistance in ohm
+x_2 = 0.003 ; // Secondary winding reactance in ohm
+// calculated data from Example 14-12
+Z_L = 0.1058 ; // Load impedance in ohm
+
+// Calculations
+alpha = V_1 / V_2 ; // Transformation ratio
+
+// case a
+R_e1 = r_1 + (alpha)^2 * r_2 ; // Equivalent internal resistance referred to the
+// primary side in ohm
+
+// case b
+X_e1 = x_1 + (alpha)^2 * x_2 ; // Equivalent internal reactance referred to the
+// primary side in ohm
+
+// case c
+Z_e1 = R_e1 + %i*(X_e1) ; // Equivalent internal impedance referred to the
+// primary side in ohm
+Z_e1_m = abs(Z_e1);//Z_e1_m=magnitude of Z_e1 in ohm
+Z_e1_a = atan(imag(Z_e1) /real(Z_e1))*180/%pi;//Z_e1_a=phase angle of Z_e1 in degrees
+
+// case d
+Z_L_prime = (alpha)^2 * (Z_L); // Equivalent secondary load impedance referred
+// to the primary side in ohm
+
+// case e
+R_L = Z_L ; // Load resistance in ohm
+X_L = 0 ; // Load reactance in ohm
+
+// Primary load current in A , when V_1 = 2300 V
+I_1 = V_1 / ( (R_e1 + alpha^2*R_L) + %i*(X_e1 + alpha^2*X_L) );
+
+// Display the results
+disp("Example 14-13 Solution : ");
+
+printf(" \n a: Equivalent internal resistance referred to the primary side :");
+printf(" \n R_c1 = %.2f ohm \n ",R_e1 );
+
+printf(" \n b: Equivalent internal reactance referred to the primary side :");
+printf(" \n X_c1 = %.2f ohm \n ",X_e1 );
+
+printf(" \n c: Equivalent internal impedance referred to the primary side : ");
+printf(" \n Z_c1 in ohm = ");disp(Z_e1);
+printf(" \n Z_c1 = %.3f <%.2f ohm \n ", Z_e1_m , Z_e1_a );
+
+printf(" \n d: Equivalent secondary load impedance referred to the primary side :");
+printf(" \n (alpha)^2 * Z_L = %.2f ohm = (alpha)^2 * R_L \n",Z_L_prime);
+
+printf(" \n e: Primary load current :\n I_1 = %f A ≈ %.f A ", I_1, I_1);
diff --git a/1092/CH14/EX14.14/Example14_14.sce b/1092/CH14/EX14.14/Example14_14.sce new file mode 100755 index 000000000..4c64e815f --- /dev/null +++ b/1092/CH14/EX14.14/Example14_14.sce @@ -0,0 +1,58 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-14
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 500 ; // kVA rating of the step-down transformer
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+R_e2 = 2 ; // Equivalent resistance referred to the
+// primary side in mΩ
+X_e2 = 6 ; // Equivalent reactance referred to the
+// primary side in mΩ
+
+// Calculations
+// case a
+I_2 = (kVA ) / V_2 ; // Rated secondary current in kA
+
+// case b
+R_e2_drop = I_2 * R_e2 ; // Full-load equivalent resistance voltage drop in volt
+
+// case c
+X_e2_drop = I_2 * X_e2 ; // Full-load equivalent reactance voltage drop in volt
+
+// case d
+// unity PF
+cos_theta2 = 1;
+sin_theta2 = sqrt(1 - (cos_theta2)^2);
+
+// Induced voltage when the transformer is delivering rated current to unity PF load
+E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 + I_2*X_e2);
+E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt
+E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees
+
+// case e
+VR = ( (E_2_m - V_2) / V_2 ) * 100 ; // Percent voltage regulation at unity PF
+
+// Display the results
+disp("Example 14-14 Solution : ");
+
+printf(" \n a: Rated secondary current :\n I_2 = %.3f kA \n ", I_2 );
+
+printf(" \n b: Full-load equivalent resistance voltage drop : ");
+printf(" \n I_2*R_c2 = %.2f V \n", R_e2_drop );
+
+printf(" \n c: Full-load equivalent reactance voltage drop : ");
+printf(" \n I_2*X_c2 = %.2f V \n", X_e2_drop );
+
+printf(" \n d: Induced voltage when the transformer is delivering rated current ");
+printf(" \n to unity PF load :\n E_2 in volt = ");disp(E_2);
+printf(" \n E_2 = %.2f <%.2f V \n ",E_2_m , E_2_a);
+
+printf(" \n e: Voltage regulation at unity PF :\n VR = %.2f percent ",VR );
diff --git a/1092/CH14/EX14.15/Example14_15.sce b/1092/CH14/EX14.15/Example14_15.sce new file mode 100755 index 000000000..b0db9121b --- /dev/null +++ b/1092/CH14/EX14.15/Example14_15.sce @@ -0,0 +1,43 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-15
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 500 ; // kVA rating of the step-down transformer
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+R_e2 = 2 ; // Equivalent resistance referred to the
+// primary side in mΩ
+X_e2 = 6 ; // Equivalent reactance referred to the
+// primary side in mΩ
+I_2 = 2.174 ; // Rated secondary current in kA
+
+cos_theta2 = 0.8 ; // lagging PF
+sin_theta2 = sqrt(1 - (cos_theta2)^2);
+
+// Calculations
+
+// case d
+// Induced voltage when the transformer is delivering rated current to 0.8 lagging PF load
+E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 + I_2*X_e2);
+E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt
+E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees
+
+// case e
+VR = ( (E_2_m - V_2) / V_2 ) * 100 ; // Percent voltage regulation at 0.8 PF lag
+
+// Display the results
+disp("Example 14-15 Solution : ");
+
+printf(" \n d: Induced voltage when the transformer is delivering rated current ");
+printf(" \n to 0.8 lagging PF load :\n E_2 in volt = ");disp(E_2);
+printf(" \n E_2 = %.2f <%.2f V \n ",E_2_m , E_2_a);
+
+printf(" \n e: Voltage regulation at 0.8 lagging PF :\n VR = %.2f percent ",VR );
+
diff --git a/1092/CH14/EX14.16/Example14_16.sce b/1092/CH14/EX14.16/Example14_16.sce new file mode 100755 index 000000000..4e3f736c4 --- /dev/null +++ b/1092/CH14/EX14.16/Example14_16.sce @@ -0,0 +1,43 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-16
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 500 ; // kVA rating of the step-down transformer
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+R_e2 = 2 ; // Equivalent resistance referred to the
+// primary side in mΩ
+X_e2 = 6 ; // Equivalent reactance referred to the
+// primary side in mΩ
+I_2 = 2.174 ; // Rated secondary current in kA
+
+cos_theta2 = 0.6 ; // leading PF
+sin_theta2 = sqrt(1 - (cos_theta2)^2);
+
+// Calculations
+
+// case d
+// Induced voltage when the transformer is delivering rated current to unity PF load
+E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 - I_2*X_e2);
+E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt
+E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees
+
+// case e
+VR = ( (E_2_m - V_2) / V_2 ) * 100 ; // Percent voltage regulation at 0.8 leading PF
+
+// Display the results
+disp("Example 14-16 Solution : ");
+
+printf(" \n d: Induced voltage when the transformer is delivering rated current ");
+printf(" \n to 0.6 leading PF load :\n E_2 in volt = ");disp(E_2);
+printf(" \n E_2 = %.2f <%.2f V \n ",E_2_m , E_2_a);
+
+printf(" \n e: Voltage regulation at 0.8 leading PF :\n VR = %.2f percent ",VR );
+
diff --git a/1092/CH14/EX14.17/Example14_17.sce b/1092/CH14/EX14.17/Example14_17.sce new file mode 100755 index 000000000..251e32a1e --- /dev/null +++ b/1092/CH14/EX14.17/Example14_17.sce @@ -0,0 +1,96 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-17
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 20 ; // kVA rating of the step-down transformer
+S = 20000 ; // power rating of the step-down transformer in VA
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+
+// w.r.t HV side following is SC-test data
+P1 = 250 ; // wattmeter reading in W
+I1 = 8.7 ; // Input current in A
+V1 = 50 ; // Input voltage in V
+
+// Calculations
+alpha = V_1 / V_2 ; // Transformation ratio
+// case a
+Z_e1 = V1 / I1 ; // Equivalent impedance w.r.t HV side in ohm
+
+R_e1 = P1 / (I1)^2 ; // Equivalent resistance w.r.t HV side in ohm
+
+theta = acosd(R_e1/Z_e1) ; // PF angle in degrees
+
+X_e1 = Z_e1*sind(theta); // Equivalent reactance w.r.t HV side in ohm
+
+// case b
+Z_e2 = Z_e1 / (alpha)^2 ; // Equivalent impedance w.r.t LV side in ohm
+
+R_e2 = R_e1 / (alpha)^2 ; // Equivalent resistance w.r.t LV side in ohm
+
+X_e2 = Z_e2*sind(theta); // Equivalent reactance w.r.t LV side in ohm
+
+// case c
+I_2 = S / V_2 ; // Rated secondary load current in A
+
+R_e2_drop = I_2 * R_e2 ; // Full-load equivalent resistance voltage drop in volt
+X_e2_drop = I_2 * X_e2 ; // Full-load equivalent reactance voltage drop in volt
+
+// At unity PF
+cos_theta2 = 1;
+sin_theta2 = sqrt(1 - (cos_theta2)^2);
+
+// Induced voltage when the transformer is delivering rated current to unity PF load
+E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 + I_2*X_e2);
+E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt
+E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees
+
+VR_unity_PF = ( (E_2_m - V_2) / V_2 ) * 100 ; // Transformer voltage regulation
+
+// case d
+// at 0.7 lagging PF
+cos_theta_2 = 0.7 ; // lagging PF
+sin_theta_2 = sqrt(1 - (cos_theta_2)^2);
+
+// Induced voltage when the transformer is delivering rated current to unity PF load
+E2 = (V_2*cos_theta_2 + I_2*R_e2) + %i*(V_2*sin_theta_2 + I_2*X_e2);
+E2_m = abs(E2);//E2_m=magnitude of E2 in volt
+E2_a = atan(imag(E2) /real(E2))*180/%pi;//E2_a=phase angle of E2 in degrees
+
+VR_lag_PF = ( (E2_m - V_2) / V_2 ) * 100 ; // Transformer voltage regulation
+
+// Display the results
+disp("Example 14-17 Solution : ");
+
+printf(" \n a: Equivalent impedance w.r.t HV side :\n Z_e1 = %.2f Ω \n",Z_e1);
+printf(" \n Equivalent resistance w.r.t HV side :\n R_e1 = %.1f Ω \n",R_e1);
+printf(" \n θ = %.f degrees \n",theta );
+printf(" \n Equivalent reactance w.r.t HV side :\n X_e1 = %.2f \n",X_e1);
+
+printf(" \n b: Equivalent impedance w.r.t LV side :");
+printf(" \n Z_e2 = %f Ω = %.2f mΩ \n",Z_e2 ,Z_e2*1000);
+printf(" \n Equivalent resistance w.r.t LV side :\n R_e2 = %f Ω \n",R_e2);
+printf(" \n R_e2 = %f Ω = %.2f mΩ \n",R_e2,R_e2*1000);
+printf(" \n Equivalent reactance w.r.t LV side :\n X_e2 = %f Ω \n",X_e2);
+printf(" \n X_e2 = %f Ω = %.2f mΩ \n",X_e2,X_e2*1000);
+
+printf(" \n c: Rated secondary load current :\n I_2 = %.f A\n",I_2);
+printf(" \n I_2*R_c2 = %.2f V \n", R_e2_drop );
+printf(" \n I_2*X_c2 = %.2f V \n", X_e2_drop );
+printf(" \n At unity PF,\n E_2 in volt = ");disp(E_2);
+printf(" \n E_2 = %.2f <%.2f V \n ",E_2_m , E_2_a);
+printf(" \n Voltage regulation at unity PF :\n VR = %.2f percent ",VR_unity_PF );
+
+printf(" \n\n d: At 0.7 lagging PF, \n E_2 in volt = ");disp(E2);
+printf(" \n E_2 = %.2f <%.2f V \n ",E2_m , E2_a);
+printf(" \n Voltage regulation at 0.7 lagging PF :\n VR = %.2f percent ",VR_lag_PF );
+
+
+
diff --git a/1092/CH14/EX14.18/Example14_18.sce b/1092/CH14/EX14.18/Example14_18.sce new file mode 100755 index 000000000..c7cd93291 --- /dev/null +++ b/1092/CH14/EX14.18/Example14_18.sce @@ -0,0 +1,26 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-18
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_sc = 50 ; // Short circuit voltage in volt
+V_1 = 2300 ; // Rated primary voltage in volt
+
+// Calculations
+P_c = poly(0,'P_c');// Making P_c as a variable just for displaying answer as per
+// textbook
+
+P_c_sc = (V_sc / V_1)^2 * P_c ; // Fraction of P_c measured by the wattmeter
+
+// Display the results
+disp("Example 14-18 Solution : ");
+
+printf(" \n Since P_c is proportional to the square of the primary voltage V_sc, ");
+printf(" \n then under short circuit conditions,the fraction of rated-core loss is :");
+printf(" \n P_c(sc) = ");disp(P_c_sc);
diff --git a/1092/CH14/EX14.19/Example14_19.sce b/1092/CH14/EX14.19/Example14_19.sce new file mode 100755 index 000000000..c039f21a5 --- /dev/null +++ b/1092/CH14/EX14.19/Example14_19.sce @@ -0,0 +1,78 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-19
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+
+kVA = 20 ; // kVA rating of the step-down transformer
+S = 20000 ; // power rating of the step-down transformer in VA
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+Z_e1 = 5.75 ; // Equivalent impedance w.r.t HV side in ohm
+R_e1 = 3.3 ; // Equivalent resistance w.r.t HV side in ohm
+X_e1 = 4.71 ; // Equivalent reactance w.r.t HV side in ohm
+
+// w.r.t HV side following is SC-test data
+P1 = 250 ; // wattmeter reading in W
+I1 = 8.7 ; // Input current in A
+V1 = 50 ; // Input voltage in V
+
+// Calculations
+// case a
+Z_e1_drop = V1 ; // High voltage impedance drop in volt
+
+// case b
+theta = acosd(R_e1/Z_e1) ; // PF angle in degrees
+
+R_e1_drop = I1*Z_e1*cosd(theta) ; //HV-side equivalent resistance voltage drop in volt
+
+// case c
+X_e1_drop = I1*Z_e1*sind(theta) ; //HV-side equivalent reactance voltage drop in volt
+
+// case d
+// At unity PF
+cos_theta1 = 1;
+sin_theta1 = sqrt(1 - (cos_theta1)^2);
+
+// Induced voltage when the transformer is delivering rated current to unity PF load
+E_1 = (V_1*cos_theta1 + I1*R_e1) + %i*(V_1*sin_theta1 + I1*X_e1);
+E_1_m = abs(E_1);//E_1_m=magnitude of E_1 in volt
+E_1_a = atan(imag(E_1) /real(E_1))*180/%pi;//E_1_a=phase angle of E_1 in degrees
+
+VR_unity_PF = ( (E_1_m - V_1) / V_1 ) * 100 ; // Transformer voltage regulation
+
+// case e
+// at 0.7 lagging PF
+cos_theta_1 = 0.7 ; // lagging PF
+sin_theta_1 = sqrt(1 - (cos_theta_1)^2);
+
+// Induced voltage when the transformer is delivering rated current to unity PF load
+E1 = (V_1*cos_theta_1 + I1*R_e1) + %i*(V_1*sin_theta_1 + I1*X_e1);
+E1_m = abs(E1);//E1_m=magnitude of E1 in volt
+E1_a = atan(imag(E1) /real(E1))*180/%pi;//E1_a=phase angle of E1 in degrees
+
+VR_lag_PF = ( (E1_m - V_1) / V_1 ) * 100 ; // Transformer voltage regulation
+
+// Display the results
+disp("Example 14-19 Solution : ");
+
+printf(" \n a: High voltage impedance drop :\n I_1*Z_e1 = V_1 = %d\n",Z_e1_drop);
+
+printf(" \n b: θ = %.f degrees \n",theta );
+printf(" \n High voltage resistance drop :\n I_1*R_e1 = %.2f \n",R_e1_drop);
+
+printf(" \n c: High voltage reactance drop :\n I_1*X_e1 = %.2f \n",X_e1_drop);
+
+printf(" \n d: At unity PF,\n E_2 in volt = ");disp(E_1);
+printf(" \n E_2 = %.2f <%.2f V \n ",E_1_m , E_1_a);
+printf(" \n Voltage regulation at unity PF :\n VR = %.2f percent ",VR_unity_PF );
+
+printf(" \n\n e: At 0.7 lagging PF, \n E_2 in volt = ");disp(E1);
+printf(" \n E_2 = %.2f <%.2f V \n ",E1_m , E1_a);
+printf(" \n Voltage regulation at 0.7 lagging PF :\n VR = %.2f percent ",VR_lag_PF );
diff --git a/1092/CH14/EX14.2/Example14_2.sce b/1092/CH14/EX14.2/Example14_2.sce new file mode 100755 index 000000000..54a22430d --- /dev/null +++ b/1092/CH14/EX14.2/Example14_2.sce @@ -0,0 +1,61 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_h = 2300 ; // high voltage in volt
+V_l = 115 ; // low voltage in volt
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 115 ; // Secondary voltage in volt
+f = 60 ; // Frequency in Hz
+S = 4.6 ; // kVA rating of the step-down transformer
+S_1 = S ;
+S_2 = S ;
+V_per_turn = 2.5 ; // Induced EMF per turn in V/turn
+// Ideal transformer
+
+// Calculations
+// case a
+N_h = V_h / V_per_turn ; // Number of high-side turns
+N_l = V_l / V_per_turn ; // Number of low-side turns
+
+N_1 = N_h;// Number of turns in the primary
+N_2 = N_l;// Number of turns in the secondary
+
+// case b
+I_1 = S_1*1000 / V_1 ; // Rated primary current in A
+I_2 = S_2*1000 / V_2 ; // Rated secondary current in A
+
+I_h = 2 ; // Rated current in A on HV side
+I_l = 40 ; // Rated current in A on LV side
+
+// case c
+// subscript c for alpha_stepdown and alpha_stepup indicates case c
+alpha_stepdown_c = N_1 / N_2 ; // step-down transformation ratio
+alpha_stepup_c = N_l / N_h ; // step-up transformation ratio
+
+// case d
+// subscript d for alpha_stepdown and alpha_stepup indicates case d
+alpha_stepdown_d = I_2 / I_1 ; // step-down transformation ratio
+alpha_stepup_d = I_h / I_l ; // step-up transformation ratio
+
+// Display the results
+disp("Example 14-2 Solution : ");
+
+printf(" \n a: Number of high-side turns :\n N_h = %d t = N_1 \n",N_h);
+printf(" \n Number of low-side turns :\n N_l = %d t = N_2\n",N_l);
+
+printf(" \n b: Rated primary current :\n I_h = I_1 = %d A \n",I_1);
+printf(" \n Rated secondary current :\n I_l = I_2 = %d A\n",I_2);
+
+printf(" \n c: step-down transformation ratio :\n α = N_1/N_2 = %d\n",alpha_stepdown_c);
+printf(" \n step-up transformation ratio :\n α = N_l/N_h = %.2f\n",alpha_stepup_c);
+
+printf(" \n d: step-down transformation ratio :\n α = I_2 / I_1 = %d\n",alpha_stepdown_d);
+printf(" \n step-up transformation ratio :\n α = I_h / I_lh = %.2f\n",alpha_stepup_d);
diff --git a/1092/CH14/EX14.20/Example14_20.sce b/1092/CH14/EX14.20/Example14_20.sce new file mode 100755 index 000000000..5fce56316 --- /dev/null +++ b/1092/CH14/EX14.20/Example14_20.sce @@ -0,0 +1,60 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-20
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 500 ; // kVA rating of the step-down transformer
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 208 ; // Secondary voltage in volt
+f = 60 ; // Frequency in Hz
+
+// SC-test data
+P_sc = 8200 ; // wattmeter reading in W
+I_sc = 217.4 ; // Short circuit current in A
+V_sc = 95 ; // Short circuit voltage in V
+
+// OC-test data
+P_oc = 1800 ; // wattmeter reading in W
+I_oc = 85 ; // Open circuit current in A
+V_oc = 208 ; // Open circuit voltage in V
+
+// Calculations
+alpha = V_1 / V_2 ; // Transformation ratio
+// case a
+P = P_sc ; // wattmeter reading in W
+I1 = I_sc ; // Short circuit current in A
+R_e1 = P / (I1)^2 ; // Equivalent resistance w.r.t HV side in ohm
+R_e2 = R_e1 / (alpha)^2 // Equivalent resistance referred to LV side in ohm
+
+// case b
+r_2 = R_e2 / 2 ; // Resistance of low-voltage side in ohm
+
+// case c
+I_m = I_oc ; // Open circuit current in A
+P_cu = (I_m)^2 * r_2 ; // Transformer copper loss of the LV side wdg during OC-test in W
+
+// case d
+P_c = P_oc - P_cu ; // Transformer core loss in W
+
+// Display the results
+disp("Example 14-20 Solution : ");
+
+printf(" \n a: Equivalent resistance w.r.t HV side :\n R_e1 = %.4f Ω\n",R_e1);
+printf(" \n Equivalent resistance w.r.t LV side :\n R_e2 = %f Ω = %.3f mΩ \n",R_e2,R_e2*1000);
+
+printf(" \n b: Resistance of LV side :\n r_2 = %f Ω = %.2f mΩ\n",r_2,r_2*1000);
+
+printf(" \n c: Transformer copper loss of the LV side wdg during OC-test : ");
+printf(" \n (I_m)^2 * r_2 = %f W \n",P_cu);
+
+printf(" \n d: Transformer core loss :\n P_c = %f W \n ",P_c);
+
+printf(" \n e: Yes.The error is approximately 5/%d = 0.278 percent,which is",P_oc);
+printf(" \n within the error produced by the instruments used in the test.");
+printf(" \n We may assume that the core loss is %d W.",P_oc);
diff --git a/1092/CH14/EX14.21/Example14_21.sce b/1092/CH14/EX14.21/Example14_21.sce new file mode 100755 index 000000000..6f6050aa4 --- /dev/null +++ b/1092/CH14/EX14.21/Example14_21.sce @@ -0,0 +1,143 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-21
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data(from Ex.14-18)
+V_sc = 50 ; // Short circuit voltage in volt
+V_1 = 2300 ; // Rated primary voltage in volt
+
+
+// Preliminary data before tabulating
+
+// from ex.14-20
+P_c = 1.8 ; // core losses in kW
+P_k = 1.8 ; // fixed losses in kW
+P_cu_rated = 8.2 ; // Rated copper loss in kW
+
+// given rating
+kVA = 500 ; // Power rating in kVA
+PF = 1 ; // power factor
+P_o = kVA * PF ; // full-load output at unity PF in kW
+
+// Calculations
+// case a
+LF1 = 1/4 ; // Load fraction
+LF2 = 1/2 ; // Load fraction
+LF3 = 3/4 ; // Load fraction
+LF4 = 5/4 ; // Load fraction
+P_cu_fl = 8.2 ; // Equivalent copper loss at full-load slip
+P_cu_LF1 = (LF1)^2 * P_cu_fl ; // Equivalent copper loss at 1/4 rated load
+P_cu_LF2 = (LF2)^2 * P_cu_fl ; // Equivalent copper loss at 1/2 rated load
+P_cu_LF3 = (LF3)^2 * P_cu_fl ; // Equivalent copper loss at 3/4 rated load
+P_cu_LF4 = (LF4)^2 * P_cu_fl ; // Equivalent copper loss at 5/4 rated load
+
+P_L_1 = P_c + P_cu_LF1 ; // Total losses in kW at 1/4 rated load
+P_L_2 = P_c + P_cu_LF2 ; // Total losses in kW at 1/2 rated load
+P_L_3 = P_c + P_cu_LF3 ; // Total losses in kW at 3/4 rated load
+P_L_fl = P_c + P_cu_fl ; // Total losses in kW at rated load
+P_L_4 = P_c + P_cu_LF4 ; // Total losses in kW at 5/4 rated load
+
+P_o_1 = P_o*LF1 ; // Total output in kW at 1/4 rated load
+P_o_2 = P_o*LF2 ; // Total output in kW at 1/2 rated load
+P_o_3 = P_o*LF3 ; // Total output in kW at 3/4 rated load
+P_o_fl = P_o ; // Total output in kW at rated load
+P_o_4 = P_o*LF4 ; // Total output in kW at 5/4 rated load
+
+P_in_1 = P_L_1 + P_o_1 ; // Total input in kW at 1/4 rated load
+P_in_2 = P_L_2 + P_o_2 ; // Total input in kW at 1/2 rated load
+P_in_3 = P_L_3 + P_o_3 ; // Total input in kW at 3/4 rated load
+P_in_fl = P_L_fl + P_o_fl ; // Total input in kW at rated load
+P_in_4 = P_L_4 + P_o_4 ; // Total input in kW at 5/4 rated load
+
+eta_1 = (P_o_1/P_in_1)*100 ; // Efficiency at 1/4 rated load
+eta_2 = (P_o_2/P_in_2)*100 ; // Efficiency at 1/2 rated load
+eta_3 = (P_o_3/P_in_3)*100 ; // Efficiency at 3/4 rated load
+eta_fl = (P_o_fl/P_in_fl)*100 ; // Efficiency at rated load
+eta_4 = (P_o_4/P_in_4)*100 ; // Efficiency at 5/4 rated load
+
+
+// case b
+PF_b = 0.8 ; // 0.8 PF lagging
+Po_1 = P_o*LF1*PF_b ; // Total output in kW at 1/4 rated load
+Po_2 = P_o*LF2*PF_b ; // Total output in kW at 1/2 rated load
+Po_3 = P_o*LF3*PF_b ; // Total output in kW at 3/4 rated load
+Po_fl = P_o*PF_b ; // Total output in kW at rated load
+Po_4 = P_o*LF4*PF_b ; // Total output in kW at 5/4 rated load
+
+Pin_1 = P_L_1 + Po_1 ; // Total input in kW at 1/4 rated load
+Pin_2 = P_L_2 + Po_2 ; // Total input in kW at 1/2 rated load
+Pin_3 = P_L_3 + Po_3 ; // Total input in kW at 3/4 rated load
+Pin_fl = P_L_fl + Po_fl ; // Total input in kW at rated load
+Pin_4 = P_L_4 + Po_4 ; // Total input in kW at 5/4 rated load
+
+eta1 = (Po_1/Pin_1)*100 ; // Efficiency at 1/4 rated load
+eta2 = (Po_2/Pin_2)*100 ; // Efficiency at 1/2 rated load
+eta3 = (Po_3/Pin_3)*100 ; // Efficiency at 3/4 rated load
+etafl = (Po_fl/Pin_fl)*100 ; // Efficiency at rated load
+eta4 = (Po_4/Pin_4)*100 ; // Efficiency at 5/4 rated load
+
+// case c
+R_e2 = 1.417e-3 ; // Equivalent resistance in ohm referred to LV side
+Pc = 1800 ; // Core losses in W
+I_2 = sqrt(Pc/R_e2); // Load current in A for max.efficiency invariant of LF
+
+// case d
+V = 208 ; // Voltage rating in volt
+I_2_rated = (kVA*1000) / V ; // Rated secondary current in A
+LF_max = I_2 / I_2_rated ; // Load fraction for max.efficiency
+
+// case e
+// subscript e for eta_max indicates case e
+cos_theta = 1;
+V_2 = V ; // secondary voltage in volt
+Pc = 1800 ; // core loss in W
+// max.efficiency for unity PF
+eta_max_e = (V_2*I_2*cos_theta) / ((V_2*I_2*cos_theta) + (Pc + I_2^2*R_e2)) * 100
+
+// case f
+// subscript f for eta_max indicates case e
+cos_theta2 = 0.8;
+// max.efficiency for 0.8 lagging PF
+eta_max_f = (V_2*I_2*cos_theta2) / ((V_2*I_2*cos_theta2) + (Pc + I_2^2*R_e2)) * 100
+
+// Display the results
+disp("Example 14-21 Solution : ");
+
+printf(" \n a: Tabulation at unity PF : ");
+printf(" \n __________________________________________________________________________________________________________");
+printf(" \n L.F \t Core loss \t Copper loss \tTotal loss \t Total Output \t Total Input \t Efficiency");
+printf(" \n \t (kW) \t (kW) \t P_L (kW) \t P_o(kW) \t P_L+P_o(kW)\t P_o/P_in(percent)");
+printf(" \n __________________________________________________________________________________________________________");
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t\t %.1f \t %.2f \t %.2f ",LF1,P_c,P_cu_LF1,P_L_1,P_o_1,P_in_1,eta_1);
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t\t %.1f \t %.2f \t %.2f ",LF2,P_c,P_cu_LF2,P_L_2,P_o_2,P_in_2,eta_2);
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t\t %.1f \t %.2f \t %.2f ",LF3,P_c,P_cu_LF3,P_L_3,P_o_3,P_in_3,eta_3);
+printf(" \n 1 \t\t %.1f \t\t %.3f \t %.3f \t %.1f \t %.2f \t %.2f ",P_c,P_cu_fl,P_L_fl,P_o_fl,P_in_fl,eta_fl);
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t %.1f \t %.2f \t %.2f ",LF4,P_c,P_cu_LF4,P_L_4,P_o_4,P_in_4,eta_4);
+printf(" \n __________________________________________________________________________________________________________\n\n");
+
+printf(" \n b: Tabulation at 0.8 PF lagging : ");
+printf(" \n __________________________________________________________________________________________________________");
+printf(" \n L.F \t Core loss \t Copper loss \tTotal loss \t Total Output \t Total Input \t Efficiency");
+printf(" \n \t (kW) \t (kW) \t P_L (kW) \t P_o(kW) \t P_L+P_o(kW)\t P_o/P_in(percent)");
+printf(" \n __________________________________________________________________________________________________________");
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t\t %.1f \t %.2f \t %.2f ",LF1,P_c,P_cu_LF1,P_L_1,Po_1,Pin_1,eta1);
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t\t %.1f \t %.2f \t %.2f ",LF2,P_c,P_cu_LF2,P_L_2,Po_2,Pin_2,eta2);
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t\t %.1f \t %.2f \t %.2f ",LF3,P_c,P_cu_LF3,P_L_3,Po_3,Pin_3,eta3);
+printf(" \n 1 \t\t %.1f \t\t %.3f \t %.3f \t %.1f \t %.2f \t %.2f ",P_c,P_cu_fl,P_L_fl,Po_fl,Pin_fl,etafl);
+printf(" \n %.2f \t %.1f \t\t %.3f \t %.3f \t %.1f \t %.2f \t %.2f ",LF4,P_c,P_cu_LF4,P_L_4,Po_4,Pin_4,eta4);
+printf(" \n __________________________________________________________________________________________________________\n\n");
+
+printf(" \n c: Load current at which max.efficiency occurs :\n I_2 = %.1f A \n",I_2);
+
+printf(" \n d: Rated load current :\n I_2(rated) = %.1f A \n",I_2_rated);
+printf(" \n Load fraction for η_max = %.3f(≃half rated load)\n ",LF_max);
+
+printf(" \n e: Max.efficiency for unity PF :\n η_max = %.2f percent \n",eta_max_e);
+
+printf(" \n f: Max.efficiency for 0.8 lagging PF :\n η_max = %.2f percent",eta_max_f);
diff --git a/1092/CH14/EX14.22/Example14_22.sce b/1092/CH14/EX14.22/Example14_22.sce new file mode 100755 index 000000000..c240921b4 --- /dev/null +++ b/1092/CH14/EX14.22/Example14_22.sce @@ -0,0 +1,74 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-22
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+P = 20 ; // Power rating of the transformer in kVA
+// Short circuit test data
+P_sc = 250 ; // Power measured in W
+V_sc = 50 ; // Short circuit voltage in volt
+I_sc = 8.7 ; // Short circuit current in A
+
+// Calculations
+// case a
+V_1b = V_1 ; // base voltage in volt
+Z_eq_pu = V_sc / V_1 ;
+
+funcprot(0) ; // Use this to avoid the message "Warning : redefining function: beta " .
+beta = acosd(P_sc/(V_sc*I_sc)); // angle in degrees
+
+Zeq_pu = Z_eq_pu*exp(%i*(beta)*(%pi/180));
+Zeq_pu_m = abs(Zeq_pu);//Zeq_pu_m=magnitude of Zeq_pu in p.u
+Zeq_pu_a = atan(imag(Zeq_pu) /real(Zeq_pu))*180/%pi;//Zeq_pu_a=phase angle of Zeq_pu in degrees
+
+// case b
+// at unity PF
+V_1_pu = 1*exp(%i*(0)*(%pi/180)) + 1*exp(%i*(0)*(%pi/180))*Z_eq_pu*exp(%i*(beta)*(%pi/180));
+// RHS is written in exponential complex form and (%pi/180) is radians to degrees conversion factor
+V_1_pu_m = abs(V_1_pu);//V_1_pu_m=magnitude of V_1_pu in volt
+V_1_pu_a = atan(imag(V_1_pu) /real(V_1_pu))*180/%pi;//V_1_pu_a=phase angle of V_1_pu in degrees
+
+// case c
+// at 0.7 PF lagging
+theta = acosd(0.7); // Power factor angle in degrees
+V1_pu = 1*exp(%i*(0)*(%pi/180)) + 1*exp(%i*(-theta)*(%pi/180))*Z_eq_pu*exp(%i*(beta)*(%pi/180));
+V1_pu_m = abs(V1_pu);//V1_pu_m=magnitude of V1_pu in volt
+V1_pu_a = atan(imag(V1_pu) /real(V1_pu))*180/%pi;//V1_pu_a=phase angle of V1_pu in degrees
+
+// case d
+VR_unity_PF = V_1_pu_m - 1 ; // voltage regulation at unity PF
+
+// case e
+VR_lag_PF = V1_pu_m - 1 ; // voltage regulation at 0.7 lagging PF
+
+// Display the results
+disp("Example 14-22 Solution : ");
+
+printf(" \n a: Z_eq(pu) = %.5f p.u \n",Z_eq_pu);
+printf(" \n β = %.f degrees \n",beta);
+printf(" \n Z_eq(pu) <β = ");disp(Zeq_pu);
+printf(" \n Z_eq(pu) <β = %.5f <%.f p.u \n ",Zeq_pu_m,Zeq_pu_a);
+
+printf(" \n b: |V_1(pu)| = ");disp(V_1_pu);
+printf(" \n |V_1(pu)| = %.4f <%.2f V \n ",V_1_pu_m , V_1_pu_a );
+
+printf(" \n c: |V_1(pu)| = ");disp(V1_pu);
+printf(" \n |V_1(pu)| = %.4f <%.2f V \n ",V1_pu_m , V1_pu_a );
+
+printf(" \n d: Voltage regulation at unity PF :\n VR = %f ",VR_unity_PF);
+printf(" \n VR = %.3f percent \n ",100*VR_unity_PF);
+
+printf(" \n e: Voltage regulation at 0.7 lagging PF :\n VR = %f ",VR_lag_PF);
+printf(" \n VR = %.2f percent \n ",100*VR_lag_PF);
+
+printf(" \n f: VRs as found by p.u method are essentially the same as those found ");
+printf(" \n in Exs.14-17 and 14-19 using the same data, for the same transformer, ");
+printf(" \n but with much less effort.");
diff --git a/1092/CH14/EX14.23/Exampl14_23.sce b/1092/CH14/EX14.23/Exampl14_23.sce new file mode 100755 index 000000000..3cf75b241 --- /dev/null +++ b/1092/CH14/EX14.23/Exampl14_23.sce @@ -0,0 +1,77 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-23
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+S = 500 ; // Power rating of the transformer in kVA
+f= 60 ; // Frequency in Hz
+
+// Open circuit test data
+V_oc = 208 ; // Open circuit voltage in volt
+I_oc = 85 ; // Open circuit current in A
+P_oc = 1800 ; // Power measured in W
+
+// Short circuit test data
+V_sc = 95 ; // Short circuit voltage in volt
+I_sc = 217.4 ; // Short circuit current in A
+P_sc = 8200 ; // Power measured in W
+
+// Calculations
+// case a
+S_b = S ; // Base voltage in kVA
+Psc = 8.2 ; // Power measured in kW during SC-test
+P_Cu_pu = Psc / S_b ; // per unit value of P_Cu at rated load
+
+// case b
+Poc = 1.8 ; // Power measured in kW during OC-test
+P_CL_pu = Poc / S_b ; // per unit value of P_CL at rated load
+
+// case c
+PF = 1 ; // unity Power factor
+eta_pu = PF / (PF + P_CL_pu + P_Cu_pu ) * 100 ; // Efficiency at rated load,unity PF
+
+// case d
+// subscript d for PF and eta_pu indicates case d
+PF_d = 0.8 ; // 0.8 lagging Power factor
+eta_pu_d = PF_d / (PF_d + P_CL_pu + P_Cu_pu ) * 100 ; // Efficiency at rated load,unity PF
+
+// case e
+LF = sqrt(P_CL_pu / P_Cu_pu); // Load fraction producing max.efficiency
+
+// case f
+eta_pu_max = (LF*PF) / ( (LF*PF) + 2*(P_CL_pu) ) * 100 ; // Maximum efficiency at unity PF load
+
+// case g
+eta_pu_max_g = (LF*PF_d) / ( (LF*PF_d) + 2*(P_CL_pu) ) * 100 ; // Maximum efficiency at 0.8 lagging PF load
+
+
+// Display the results
+disp("Example 14-23 Solution : ");
+
+printf(" \n a: Per unit copper loss at rated load :");
+printf(" \n P_Cu(pu) = %.4f p.u = R_eq(pu)\n",P_Cu_pu);
+
+printf(" \n a: Per unit core loss at rated load :");
+printf(" \n P_CL(pu) = %.4f p.u \n",P_CL_pu);
+
+printf(" \n c: Efficiency at rated load,unity PF :\n η_pu = %.2f percent \n",eta_pu);
+
+printf(" \n c: Efficiency at rated load,0.8 lagging PF :\n η_pu = %.2f percent \n",eta_pu_d);
+
+printf(" \n e: Load fraction producing max.efficiency :\n L.F = %.3f \n ",LF );
+
+printf(" \n f: Maximum efficiency at unity PF load :\n η_pu(max) = %.2f percent \n",eta_pu_max);
+
+printf(" \n g: Maximum efficiency at 0.8 lagging PF load :\n η_pu(max) = %.2f percent \n",eta_pu_max_g);
+
+printf(" \n h: All efficiency values are identical to those computed in solution to Ex.14-21. \n");
+
+printf(" \n i: Per-unit method is much simpler and less subject to error than conventional method.");
diff --git a/1092/CH14/EX14.24/Example14_24.sce b/1092/CH14/EX14.24/Example14_24.sce new file mode 100755 index 000000000..8b0f39e27 --- /dev/null +++ b/1092/CH14/EX14.24/Example14_24.sce @@ -0,0 +1,60 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-24
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data(From Ex.14-23)
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+S = 500 ; // Power rating of the transformer in kVA
+f= 60 ; // Frequency in Hz
+
+// Open circuit test data
+V_oc = 208 ; // Open circuit voltage in volt
+I_oc = 85 ; // Open circuit current in A
+P_oc = 1800 ; // Power measured in W
+
+// Short circuit test data
+V_sc = 95 ; // Short circuit voltage in volt
+I_sc = 217.4 ; // Short circuit current in A
+P_sc = 8200 ; // Power measured in W
+
+// Calculations
+// Preliminary calculations
+S_b = S ; // Base voltage in kVA
+Psc = 8.2 ; // Power measured in kW during SC-test
+P_Cu_pu = Psc / S_b ; // per unit value of P_Cu at rated load
+
+Poc = 1.8 ; // Power measured in kW during OC-test
+P_CL_pu = Poc / S_b ; // per unit value of P_CL at rated load
+
+// case a
+LF1 = 3/4 ; // Load fraction of rated load
+PF1 = 1 ; // unity Power factor
+eta_pu_LF1 = (LF1*PF1) / ((LF1*PF1) + P_CL_pu + (LF1)^2*P_Cu_pu ) * 100 ; // Efficiency at rated load,unity PF
+
+// case b
+LF2 = 1/4 ; // Load fraction of rated load
+PF2 = 0.8 ; // 0.8 lagging PF
+eta_pu_LF2 = (LF2*PF2) / ((LF2*PF2) + P_CL_pu + (LF2)^2*P_Cu_pu ) * 100 ; // Efficiency at 1/4 rated load,0.8 lagging PF
+
+// case c
+LF3 = 5/4 ; // Load fraction of rated load
+PF3 = 0.8 ; // 0.8 leading PF
+eta_pu_LF3 = (LF3*PF3) / ((LF3*PF3) + P_CL_pu + (LF3)^2*P_Cu_pu ) * 100 ; // Efficiency at r1/4 rated load,0.8 leading PF
+
+
+// Display the results
+disp("Example 14-24 Solution : ");
+
+printf(" \n Efficiency(pu) :\n ");
+printf(" \n a: η_pu at %.2f rated-load = %.2f percent \n",LF1,eta_pu_LF1);
+
+printf(" \n b: η_pu at %.2f rated-load = %.2f percent \n",LF2,eta_pu_LF2);
+
+printf(" \n c: η_pu at %.2f rated-load = %.2f percent \n",LF3,eta_pu_LF3);
diff --git a/1092/CH14/EX14.25/Example14_25.sce b/1092/CH14/EX14.25/Example14_25.sce new file mode 100755 index 000000000..1af6dc0ff --- /dev/null +++ b/1092/CH14/EX14.25/Example14_25.sce @@ -0,0 +1,87 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-25
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA_1 = 500 ; // Power rating of the transformer 1 in kVA
+R_1_pu = 0.01 ; // per-unit value of resistance of the transformer 1
+X_1_pu = 0.05 ; // per-unit value of reactance of the transformer 1
+Z_1_pu = R_1_pu + %i*X_1_pu ; //per-unit value of impedance of the transformer 1
+
+PF = 0.8 ; // lagging power factor
+V_2 = 400 ; // Secondary voltage in volt
+S_load = 750 ; // Increased system load in kVA
+
+kVA_2 = 250 ; // Power rating of the transformer 2 in kVA
+R_pu_2 = 0.015 ; // per-unit value of resistance of the transformer 2
+X_pu_2 = 0.04 ; // per-unit value of reactance of the transformer 2
+
+// smaller transformer secondary voltageis same as larger transformer
+
+// Calculations
+// Preliminary calculations
+Z_pu_1 = R_pu_2 + %i*X_pu_2 ; // New transformer p.u. impedance
+
+// Calculations
+// case a
+V_b1 = 400 ; // base voltage in volt
+V_b2 = 400 ; // base voltage in volt
+Z_pu_2 = (kVA_1/kVA_2)*(V_b1/V_b2)^2 * (Z_pu_1); // New transformer p.u impedance
+Z_2_pu = Z_pu_2 ; //New transformer p.u impedance
+
+// case b
+cos_theta = PF ; // Power factor
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+S_t_conjugate = (kVA_1 + kVA_2)*(cos_theta + %i*sin_theta); // kVA of total load
+
+// case c
+S_2_conjugate = S_t_conjugate * ( Z_1_pu /(Z_1_pu + Z_2_pu) ); // Portion of load carried by the smaller transformer in kVA
+S_2_conjugate_m = abs(S_2_conjugate);//S_2_conjugate_m=magnitude of S_2_conjugate in kVA
+S_2_conjugate_a = atan(imag(S_2_conjugate) /real(S_2_conjugate))*180/%pi;//S_2_conjugate_a=phase angle of S_2_conjugate in degrees
+
+// case d
+S_1_conjugate = S_t_conjugate * ( Z_2_pu/(Z_1_pu + Z_2_pu) ); // Portion of load carried by the original transformer in kVA
+S_1_conjugate_m = abs(S_1_conjugate);//S_1_conjugate_m=magnitude of S_1_conjugate in kVA
+S_1_conjugate_a = atan(imag(S_1_conjugate) /real(S_1_conjugate))*180/%pi;//S_1_conjugate_a=phase angle of S_1_conjugate in degrees
+
+// case e
+S_1 = S_1_conjugate_m ;
+S_b1 = kVA_1 ; // base power in kVA of trancsformer 1
+LF1 = (S_1 / S_b1)*100 ; // Load fraction of the original transformer in percent
+
+// case f
+S_2 = S_2_conjugate_m ;
+S_b2 = kVA_2 ; // base power in kVA of trancsformer 2
+LF2 = (S_2 / S_b2)*100 ; // Load fraction of the original transformer in percent
+
+// Display the results
+disp("Example 14-25 Solution : ");
+
+printf(" \n a: New transformer p.u impedance :\n Z_p.u.2 in p.u = ");disp(Z_pu_2);
+
+printf(" \n b: kVA of total load :\n S*_t in kVA = ");disp(S_t_conjugate);
+
+printf(" \n c: Portion of load carried by the smaller transformer in kVA :");
+printf(" \n S*_2 in kVA = ");disp(S_2_conjugate);
+printf(" \n S*_2 = %.1f <%.2f kVA (inductive load)\n",S_2_conjugate_m,S_2_conjugate_a);
+
+printf(" \n d: Portion of load carried by the original transformer in kVA :");
+printf(" \n S*_2 in kVA = ");disp(S_1_conjugate);
+printf(" \n S*_2 = %.1f <%.2f kVA (inductive load)\n",S_1_conjugate_m,S_1_conjugate_a);
+
+printf(" \n e: Load fraction of the original transformer :\n L.F.1 = %.1f percent\n",LF1);
+
+printf(" \n f: Load fraction of the original transformer :\n L.F.2 = %.1f percent\n",LF2);
+
+printf(" \n g: Yes. Reduce the no-load voltage of the new transformer to some value ");
+printf(" \n below that of its present value so that its share of the load is reduced.");
+
+
+
+
diff --git a/1092/CH14/EX14.26/Example14_26.sce b/1092/CH14/EX14.26/Example14_26.sce new file mode 100755 index 000000000..5de5d722c --- /dev/null +++ b/1092/CH14/EX14.26/Example14_26.sce @@ -0,0 +1,110 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-26
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data(From Ex.14-25)
+kVA_1 = 500 ; // Power rating of the transformer 1 in kVA
+R_1_pu = 0.01 ; // per-unit value of resistance of the transformer 1
+X_1_pu = 0.05 ; // per-unit value of reactance of the transformer 1
+Z_1_pu = R_1_pu + %i*X_1_pu ; //per-unit value of impedance of the transformer 1
+
+PF = 0.8 ; // lagging power factor
+V = 400 ; // Secondary voltage in volt
+S_load = 750 ; // Increased system load in kVA
+
+kVA_2 = 250 ; // Power rating of the transformer 2 in kVA
+R_pu_2 = 0.015 ; // per-unit value of resistance of the transformer 2
+X_pu_2 = 0.04 ; // per-unit value of reactance of the transformer 2
+
+// smaller transformer secondary voltageis same as larger transformer
+
+// Calculations
+// Preliminary calculations
+Z_pu_1 = R_pu_2 + %i*X_pu_2 ; // New transformer p.u. impedance
+
+// case a
+V_b = V ; // (given)
+
+//case b
+S_b =500*1000 ; // base power in VA
+I_b = S_b / V_b ; // base current in A
+
+// case c
+Z_b = V^2/S_b ; // Base impedance in ohm
+
+// case d
+Z_1 = Z_b * Z_1_pu * 1000 ; // Actual impedance of larger transformer in milli-ohm
+Z_1_m = abs(Z_1);//Z_1_m=magnitude of Z_1 in ohm
+Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees
+
+// case e
+V_b1 = V_b ; // base voltage in volt
+V_b2 = V_b ; // base voltage in volt
+Z_pu_2 = (kVA_1/kVA_2)*(V_b1/V_b2)^2 * (Z_pu_1); // New transformer p.u impedance
+Z_2_pu = Z_pu_2 ; //New transformer p.u impedance
+
+Z_2 = Z_b * Z_2_pu*1000 ; // Actual impedance of smaller transformer in milli-ohm
+Z_2_m = abs(Z_2);//Z_2_m=magnitude of Z_2 in ohm
+Z_2_a = atan(imag(Z_2) /real(Z_2))*180/%pi;//Z_2_a=phase angle of Z_2 in degrees
+
+// case f
+cos_theta = 0.8 ; // Power factor
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+S_T = (kVA_1 + kVA_2)*(cos_theta - %i*sin_theta); // kVA of total load
+
+I_T = S_T*1000 / V_b ; // Total current in A
+
+I_1 = I_T*(Z_2/(Z_1 + Z_2)); // Actual current delivered by larger transformer in A
+I_1_m = abs(I_1);//I_1_m=magnitude of I_1 in A
+I_1_a = atan(imag(I_1) /real(I_1))*180/%pi;//I_1_a=phase angle of I_1 in degrees
+
+// case g
+I_2 = I_T*(Z_1/(Z_1 + Z_2)); // Actual current delivered by larger transformer in A
+I_2_m = abs(I_2);//I_2_m=magnitude of I_2 in A
+I_2_a = atan(imag(I_2) /real(I_2))*180/%pi;//I_2_a=phase angle of I_2 in degrees
+
+// case h
+Z1 = Z_1/1000 ; // Z_1 in ohm
+E_1 = I_1*Z1 + V_b ; // No-load voltage of larger Tr. in volt
+E_1_m = abs(E_1);//E_1_m=magnitude of E_1 in volt
+E_1_a = atan(imag(E_1) /real(E_1))*180/%pi;//E_1_a=phase angle of E_1 in degrees
+
+
+// case i
+Z2 = Z_2/1000 ; // Z_2 in ohm
+E_2 = I_2*Z2 + V_b ; // No-load voltage of smaller Tr. in volt
+E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt
+E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees
+
+// Display the results
+disp("Example 14-26 Solution : ");
+
+printf(" \n a: Base voltage :\n V_b = %d <0 V (given)\n",V_b);
+
+printf(" \n b: Base current :\n I_b = %.2f A \n",I_b);
+
+printf(" \n c: Base impedance :\n Z_b = %.2f ohm\n",Z_b);
+
+printf(" \n d: Actual impedance of larger transformer :\n Z_1 in mΩ = \n");disp(Z_1);
+printf(" \n Z_1 = %.2f <%.2f mΩ \n ",Z_1_m,Z_1_a);
+
+printf(" \n e: Actual impedance of smaller transformer :\n Z_1 in mΩ = \n");disp(Z_2);
+printf(" \n Z_1 = %.2f <%.2f mΩ \n ",Z_2_m,Z_2_a);
+
+printf(" \n f: Actual current delivered by larger transformer :\n I_1 in A = ");disp(I_1);
+printf(" \n I_1 = %.2f <%.2f A \n ",I_1_m,I_1_a);
+
+printf(" \n g: Actual current delivered by smaller transformer :\n I_2 in A = ");disp(I_2);
+printf(" \n I_1 = %.2f <%.2f A \n ",I_2_m,I_2_a);
+
+printf(" \n h: No-load voltage of larger Tr :\n E_1 in volt = ");disp(E_1);
+printf(" \n E_1 = %.2f <%.2f V \n ",E_1_m,E_1_a);
+
+printf(" \n i: No-load voltage of smaller Tr :\n E_2 in volt = ");disp(E_2);
+printf(" \n E_1 = %.2f <%.2f V \n ",E_2_m,E_2_a);
diff --git a/1092/CH14/EX14.27/Example14_27.sce b/1092/CH14/EX14.27/Example14_27.sce new file mode 100755 index 000000000..9beee3d5d --- /dev/null +++ b/1092/CH14/EX14.27/Example14_27.sce @@ -0,0 +1,102 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-27
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// From diagram in fig.14-23a
+P_L = 14400 ; // Load output power in W
+V_L = 120 ; // Load voltage in volt
+V_b1 = 120 ; // base voltage at point 1 in volt
+V_b2 = 600 ; // base voltage at point 2 in volt
+V_b3 = 120 ; // base voltage at point 3 in volt
+S_b3 = 14.4 ; // base power in kVA
+X_2 = %i*0.25 ; // reactance in p.u
+X_1 = %i*0.2 ; // reactance in p.u
+I_L = 120 ; // Load current in A
+
+// Calculations
+// case a
+R_L = P_L / (V_L^2); // Resistance of the load in ohm
+
+// case b
+Z_bL = (V_b3^2)/(S_b3*1000); // Base impedance in ohm
+
+// case c
+Z_L_pu = R_L / Z_bL ; // per unit load impedance
+
+// case d
+Z_2_pu = X_2 ; // per unit impedance of Tr.2
+
+// case e
+Z_1_pu = X_1 ; // per unit impedance of Tr.1
+
+// case g
+I_bL = (S_b3*1000)/V_b3 ; // Base current in load in A
+
+// case h
+I_L_pu = I_L / I_bL ; // per unit load current
+
+// case i
+V_R_pu = I_L_pu * Z_L_pu ; // per unit voltage across load
+
+// case j
+I_S_pu = I_L_pu ; //per unit current of source
+Z_T_pu = Z_L_pu + Z_1_pu + Z_2_pu ; // Total p.u impedance
+V_S_pu = I_S_pu * Z_T_pu ; // per unit voltage of source
+V_S_pu_m = abs(V_S_pu);//V_S_pu_m=magnitude of V_S_pu in p.u
+V_S_pu_a = atan(imag(V_S_pu) /real(V_S_pu))*180/%pi;//V_S_pu_a=phase angle of V_S_pu in degrees
+
+// case k
+V_S = V_S_pu * V_b1 ; // Actual voltage across source in volt
+V_S_m = abs(V_S);//V_S_m=magnitude of V_S in volt
+V_S_a = atan(imag(V_S) /real(V_S))*180/%pi;//V_S_a=phase angle of V_S in degrees
+
+
+// case l
+I_x_pu = I_L_pu ; // p.u current at point x
+Z_x_pu = Z_L_pu + Z_2_pu ; // p.u impedance at point x
+V_x_pu = I_x_pu * Z_x_pu ; // p.u voltage at point x
+
+// case m
+V_x = V_x_pu * V_b2 ; // Actual voltage at point x in volt
+V_x_m = abs(V_x);//V_x_m=magnitude of V_x in volt
+V_x_a = atan(imag(V_x) /real(V_x))*180/%pi;//V_x_a=phase angle of V_x in degrees
+
+
+// Display the results
+disp("Example 14-27 Solution : ");
+
+printf(" \n a: Resistance of the load :\n R_L = %d Ω \n",R_L);
+
+printf(" \n b: Base impedance :\n Z_bL = %d Ω \n",Z_bL);
+
+printf(" \n c: per unit load impedance :\n Z_L(pu) = ");disp(Z_L_pu);
+
+printf(" \n d: per unit impedance of Tr.2 :\n Z_2(pu) = ");disp(Z_2_pu);
+
+printf(" \n e: per unit impedance of Tr.1 :\n Z_1(pu) = ");disp(Z_1_pu);
+
+printf(" \n f: See Fig.14-23b \n");
+
+printf(" \n g: Base current in load :\n I_bL = %d A (resistive)\n",I_bL);
+
+printf(" \n h: per unit load current :\n I_L_pu = ");disp(I_L_pu);
+
+printf(" \n i: per unit voltage across load :\n V_R_pu");disp(V_R_pu);
+
+printf(" \n j: per unit voltage of source :\n V_S_pu = ");disp(V_S_pu);
+printf(" \n V_S_pu = %.3f <%.2f p.u \n",V_S_pu_m,V_S_pu_a);
+
+printf(" \n k: Actual voltage across source :\n V_S in volt = ");disp(V_S);
+printf(" \n V_S = %.1f <%.2f V \n",V_S_m,V_S_a);
+
+printf(" \n l: p.u voltage at point x :\n V_x(pu) = ");disp(V_x_pu);
+
+printf(" \n m: Actual voltage at point x :\n V_x in volt = ");disp(V_x);
+printf(" \n V_S = %.1f <%.2f V \n",V_x_m,V_x_a);
diff --git a/1092/CH14/EX14.28/Example14_28.sce b/1092/CH14/EX14.28/Example14_28.sce new file mode 100755 index 000000000..ff7e7c293 --- /dev/null +++ b/1092/CH14/EX14.28/Example14_28.sce @@ -0,0 +1,99 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-28
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// From diagram in fig.14-24a
+V_1 = 11 ; // Tr.1 voltage in kV
+V_b1 = 11 ; // Base Tr.1 voltage in kV
+S_1 = 50 ; // KVA rating of power for Tr.1
+S_2 = 100 ; // KVA rating of power for Tr.2
+Z_1_pu = %i*0.1 ; // per unit impedance of Tr.1
+Z_2_pu = %i*0.1 ; // per unit impedance of Tr.2
+V_b2 = 55 ; // Base Tr.2 voltage in kV
+S_b = 100 ; // base power in kVA
+PF = 0.8 ; // power factor of the Tr.s
+
+Z_line = %i*200 ; // line impedance in ohm
+
+V_L = 10 ;// Load voltage in kV
+V_Lb3 = 11 ; // base line voltage at point 3
+
+V_b3 = 11 ; // line voltage at point 3
+
+P_L = 50 ; // Power rating of each Tr.s in kW
+cos_theta_L = 0.8 ; // PF operation of each Tr.s
+
+// Calculations
+// case a
+Z_T1 = Z_1_pu * (V_1/V_b1)^2 * (S_2/S_1); // p.u impedance of Tr.1
+
+// case b
+Z_T2 = Z_2_pu * (V_1/V_b3)^2 * (S_2/S_1); // p.u impedance of Tr.1
+
+// case c
+V_b = 55 ; // base voltage in volt
+Z_b_line = (V_b^2)/S_b * 1000 ; // base line impedance in ohm
+Z_line_pu = Z_line / Z_b_line ; // p.u impedance of the transmission line
+
+// case d
+V_L_pu = V_L / V_Lb3 ; // p.u voltage across load
+
+// case e
+//See Fig.14-24b
+
+// case f
+I_bL = S_b / V_b3 ; // base current in load in A
+
+// case g
+VL = 11 ; // load voltage in kV
+cos_theta_L = 0.8 ; // power factor
+I_L = P_L / (VL*cos_theta_L);
+I_L_pu = I_L / I_bL ; // p.u load current
+theta = acosd(0.8);
+I_Lpu = I_L_pu*(cosd(theta) - %i*sind(theta)) ;// p.u current in complex form
+
+// case h
+Z_series_pu = Z_T1 + Z_line_pu + Z_T2 ; // p.u series impedance os the transmission line
+V_S_pu = I_Lpu * Z_series_pu + V_L_pu ; // p.u source voltage
+V_S_pu_m = abs(V_S_pu);//V_S_pu_m=magnitude of V_S_pu in p.u
+V_S_pu_a = atan(imag(V_S_pu) /real(V_S_pu))*180/%pi;//V_S_pu_a=phase angle of V_S_pu in degrees
+
+// case i
+V_S = V_S_pu_m * V_b1 ; // Actual value of source voltage in kV
+V_source = V_S*exp(%i*(V_S_pu_a)*(%pi/180)); // V_S in exponential form
+V_source_m = abs(V_source);//V_source_m=magnitude of V_source in p.u
+V_source_a = atan(imag(V_source) /real(V_source))*180/%pi;//V_source_a=phase angle of V_source in degrees
+
+
+// Display the results
+disp("Example 14-28 Solution : ");
+
+printf(" \n a: p.u impedance of Tr.1 :\n Z_T1 = ");disp(Z_T1);
+
+printf(" \n b: p.u impedance of Tr.2 :\n Z_T2 = ");disp(Z_T2);
+
+printf(" \n c: base line impedance in ohm :\n Z_b(line) = %d ohm \n",Z_b_line);
+printf(" \n p.u impedance of the transmission line :\n Z(line)_pu = ");disp(Z_line_pu);
+
+printf(" \n d: p.u voltage across load :\n V_L_pu = ");disp(V_L_pu);
+
+printf(" \n e: See Fig.14-24b \n");
+
+printf(" \n f: base current in load :\n I_bL = %.3f A \n",I_bL);
+
+printf(" \n g: Load current :\n I_L = %f A \n",I_L);
+printf(" \n p.u load current:\n I_L_pu = %.3f at %.1f PF lagging \n",I_L_pu,PF);
+printf(" \n p.u current in complex form :\n I_L_pu = ");disp(I_Lpu);
+
+printf(" \n h: per unit voltage of source :\n V_S_pu = ");disp(V_S_pu);
+printf(" \n V_S_pu = %.3f <%.2f p.u \n",V_S_pu_m,V_S_pu_a);
+
+printf(" \n i: Actual voltage across source :\n V_S in kV = ");disp(V_source);
+printf(" \n V_S = %.1f <%.2f kV \n",V_source_m,V_source_a);
diff --git a/1092/CH14/EX14.29/Example14_29.sce b/1092/CH14/EX14.29/Example14_29.sce new file mode 100755 index 000000000..74be61aed --- /dev/null +++ b/1092/CH14/EX14.29/Example14_29.sce @@ -0,0 +1,73 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-29
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// From diagram in fig.14-25a
+Z_pu_1 = %i*0.1 ; // p.u impedance
+MVA_2 = 80 ; // MVA rating os system 2
+MVA_1 = 100 ; // MVA rating of Tr.s 1 and 2
+V_2 = 30 ; // voltage in KV
+V_1 = 32 ; // voltage in KV
+
+Z_pu_2 = %i*0.15 ; // p.u impedance
+
+V_b1 = 100 ; // base voltage of Tr.1
+
+Z_line = %i*60 ; // Line impedance
+
+MVA_M1 = 20 ; // MVA rating of motor load 1
+Z_pu_M1 = %i*0.15 ; // p.u impedance of motor load M1
+
+MVA_M2 = 35 ; // MVA rating of motor load 2
+Z_pu_M2 = %i*0.25 ; // p.u impedance of motor load M2
+
+MVA_M3 = 25 ; // MVA rating of motor load 3
+Z_pu_M3 = %i*0.2 ; // p.u impedance of motor load M3
+
+V_M = 28 ; // voltage across motor loads M1,M2,M3 in kV
+
+// Calculations
+// case a
+Z_1_pu = Z_pu_1*(MVA_2/MVA_1)*(V_2/V_1)^2 ; // p.u imepedance of T1
+
+// case b
+Z_2_pu = Z_pu_2*(MVA_2/MVA_1)*(V_2/V_1)^2 ; // p.u imepedance of T2
+
+// case c
+V_b_line = V_b1*(V_1/V_2) ; // base voltage of the long-transmission line in kV
+
+// case d
+MVA_b = 80 ; // MVA rating
+V_b = V_b_line ;
+Z_line_pu = Z_line*(MVA_b/(V_b)^2); // p.u impedance of the transmission line
+
+// case e
+Z_M1_pu = Z_pu_M1 * (MVA_2/MVA_M1)*(V_M/V_1)^2 ; // p.u impedance of motor load M1
+Z_M2_pu = Z_pu_M2 * (MVA_2/MVA_M2)*(V_M/V_1)^2 ; // p.u impedance of motor load M2
+Z_M3_pu = Z_pu_M3 * (MVA_2/MVA_M3)*(V_M/V_1)^2 ; // p.u impedance of motor load M3
+
+// Display the results
+disp("Example 14-29 Solution : ");
+
+printf(" \n a: p.u imepedance of T1 :\n Z_1(pu) = ");disp(Z_1_pu);
+
+printf(" \n b: p.u imepedance of T2 :\n Z_2(pu) = ");disp(Z_2_pu);
+
+printf(" \n c: base voltage of the long-transmission line :\n V_b(line) = %.1f kV \n",V_b_line);
+
+printf(" \n d: p.u impedance of the transmission line :\n Z(line)_pu = ");disp(Z_line_pu);
+
+printf(" \n e: p.u impedance of motor load M1 :\n Z_M1(pu) = ");disp(Z_M1_pu);
+
+printf(" \n f: p.u impedance of motor load M1 :\n Z_M2(pu) = ");disp(Z_M2_pu);
+
+printf(" \n g: p.u impedance of motor load M1 :\n Z_M3(pu) = ");disp(Z_M3_pu);
+
+printf(" \n h: See Fig.14-25b.");
diff --git a/1092/CH14/EX14.3/Example14_3.sce b/1092/CH14/EX14.3/Example14_3.sce new file mode 100755 index 000000000..aae33ec27 --- /dev/null +++ b/1092/CH14/EX14.3/Example14_3.sce @@ -0,0 +1,34 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+N_1 = 500 ; // Number of primary turns in the audio output transformer
+N_2 = 25 ; // Number of secondary turns in the audio output transformer
+Z_L = 8 ; // Speaker impedance in ohm
+V_1 = 10 ; // Output voltage of the audio output transformer in volt
+
+// Calculations
+// case a
+alpha = N_1/N_2 ; // step-down transformation ratio
+Z_1 = (alpha)^2 * Z_L ; // Impedance reflected to the transformer primary
+// at the output of Tr in ohm
+
+// case b
+I_1 = V_1 / Z_1 ; // Primary current in A
+
+// Display the results
+disp("Example 14-3 Solution : ");
+
+printf(" \n a: Transformation ratio :\n α = %d\n",alpha);
+printf(" \n Impedance reflected to the transformer primary at the output of Tr :");
+printf(" \n Z_1 = %d ohm \n ",Z_1);
+
+printf(" \n b: Matching transformer primary current :\n I_1 = %f A",I_1);
+printf(" \n I_1 = %.3f mA ",1000 * I_1);
diff --git a/1092/CH14/EX14.30/Example14_30.sce b/1092/CH14/EX14.30/Example14_30.sce new file mode 100755 index 000000000..0a842586c --- /dev/null +++ b/1092/CH14/EX14.30/Example14_30.sce @@ -0,0 +1,55 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-30
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// subscripts a,b,c for the current, voltages indicates respective cases a ,b ,c.
+// from fig.14-27a
+V_pa = 1000 ; // Phase voltage in volt
+I_1a = 1 ; // line current in primary in A
+V_2a = 100 ; // voltage across secondary in V
+Ic_a = 10 ; // current in lower half of auto-transformer in A
+
+// from fig.14-26b
+V_s = 100 ; // voltage in secondary wdg in V
+I_2b = 10 ; // current in secondary in A
+V_1b = 1000 ; // voltage across primary in V
+Ic_b = 1 ; // current in lower half of auto-transformer in A
+
+// Calculations
+// case a
+S_T1 = (V_pa*I_1a + V_2a*I_1a)/1000 ; // Total kVA transfer in step-down mode
+
+// case b
+S_T2 = (V_s*I_2b + V_1b*I_2b)/1000 ; // Total kVA transfer in step-up mode
+
+// case c
+S_x_former_c = V_pa*I_1a/1000 ; // kVA rating of th autotransformer in Fig.14-27a
+
+// case d
+V_1 = V_pa ;
+S_x_former_d = V_1*Ic_b/1000 ; // kVA rating of th autotransformer in Fig.14-26b
+
+
+// Display the results
+disp("Example 14-30 Solution : ");
+
+printf(" \n a: Total kVA transfer in step-down mode :\n S_T = %.1f kVA transferred \n",S_T1);
+
+printf(" \n b: Total kVA transfer in step-up mode :\n S_T = %.1f kVA transferred \n",S_T2);
+
+printf(" \n c: kVA rating of th autotransformer in Fig.14-27a:\n S_x-former = %d kVA \n ",S_x_former_c);
+
+printf(" \n d: kVA rating of th autotransformer in Fig.14-26b:\n S_x-former = %d kVA \n ",S_x_former_d);
+
+printf(" \n e: Both transformers have the same kVA rating of 1 kVA since the same ");
+printf(" \n autotransformer is used in both parts.Both transformers transform ");
+printf(" \n a total of 1 KVA. But the step-down transformer in part(a) conducts ");
+printf(" \n only 0.1 kVA while the step-up transformer in the part(b) conducts 10");
+printf(" \n kVA from the primary to the secondary.");
diff --git a/1092/CH14/EX14.31/Example14_31.sce b/1092/CH14/EX14.31/Example14_31.sce new file mode 100755 index 000000000..dd4584bcb --- /dev/null +++ b/1092/CH14/EX14.31/Example14_31.sce @@ -0,0 +1,130 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-31
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S = 500 ; // kVA rating of distribution transformer
+// given data from ex.14-20
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 208 ; // Secondary voltage in volt
+f = 60 ; // Frequency in Hz
+
+// SC-test data
+P_sc = 8200 ; // wattmeter reading in W
+I_sc = 217.4 ; // Short circuit current in A
+V_sc = 95 ; // Short circuit voltage in V
+
+// OC-test data
+P_oc = 1800 ; // wattmeter reading in W
+I_oc = 85 ; // Open circuit current in A
+V_oc = 208 ; // Open circuit voltage in V
+
+LF_1 = 20 ; // Load fraction in percent
+LF_2 = 40 ; // Load fraction in percent
+LF_3 = 80 ; // Load fraction in percent
+LF_fl = 100 ; // rated load in percent
+LF_4 = 125 ; // Load fraction in percent
+
+LF1 = 0.2 ; // Load fraction
+LF2 = 0.4 ; // Load fraction
+LF3 = 0.8 ; // Load fraction
+LF4 = 1.25 ; // Load fraction
+
+PF1 = 0.7 ; // power factor
+PF2 = 0.8 ; // power factor
+PF3 = 0.9 ; // power factor
+PF_fl = 1 ; // power factor
+PF4 = 0.85 ; // power factor
+
+t1 = 4 ; // period of operation in hours
+t2 = 4 ; // period of operation in hours
+t3 = 6 ; // period of operation in hours
+t_fl = 6 ; // period of operation in hours
+t4 = 2 ; // period of operation in hours
+
+// Calculations
+// case a
+t = 24 ; // hrs in a day
+P_c = P_oc ; // wattmeter reading in W (OC test)
+W_c = (P_c * t)/1000 ; // COre loss over 24 hour period
+
+// case b
+Psc = P_sc/1000 ; // wattmeter reading in W (SC test)
+P_loss_1 = (LF1^2)*Psc ; // Power loss in kW for 20% Load
+P_loss_2 = (LF2^2)*Psc ; // Power loss in kW for 40% Load
+P_loss_3 = (LF3^2)*Psc ; // Power loss in kW for 80% Load
+P_loss_fl = Psc ; // Power loss in kW for 100% Load
+P_loss_4 = (LF4^2)*Psc ; // Power loss in kW for 125% Load
+
+// energy loss in kWh
+energy_loss1 = P_loss_1 * t1 ; // Enegry loss in kWh for 20% Load
+energy_loss2 = P_loss_2 * t2 ; // Enegry loss in kWh for 40% Load
+energy_loss3 = P_loss_3 * t3 ; // Enegry loss in kWh for 80% Load
+energy_loss_fl = P_loss_fl * t_fl ; // Enegry loss in kWh for 100% Load
+energy_loss4 = P_loss_4 * t4 ; // Enegry loss in kWh for 125% Load
+
+// Total energy losses in 24hrs
+W_loss_total = energy_loss1 + energy_loss2 + energy_loss3 + energy_loss_fl + energy_loss4 ;
+
+// case c
+P_1 = LF1*S*PF1 ; // Power output for 20% load
+P_2 = LF2*S*PF2 ; // Power output for 40% load
+P_3 = LF3*S*PF3 ; // Power output for 80% load
+P_fl = S*PF_fl ; // Power output for 100% load
+P_4 = LF4*S*PF4 ; // Power output for 125% load
+
+Energy_1 = P_1*t1 ; // Energy delivered in kWh for 20%load
+Energy_2 = P_2*t2 ; // Energy delivered in kWh for 40%load
+Energy_3 = P_3*t3 ; // Energy delivered in kWh for 80%load
+Energy_fl = P_fl*t_fl ; // Energy delivered in kWh for 100%load
+Energy_4 = P_4*t4 ; // Energy delivered in kWh for 125%load
+
+// Total energy delivered in 24hrs
+W_out_total = Energy_1 + Energy_2 + Energy_3 + Energy_fl + Energy_4 ;
+
+// case d
+eta = W_out_total / (W_out_total + W_c + W_loss_total) * 100 ; // All-day efficiency
+
+// Display the results
+disp("Example 14-31 Solution : ");
+
+printf(" \n a: Total energy core loss for 24hrs, including 2hours at no-load,");
+printf(" \n W_c = %.1f kWh \n ",W_c);
+
+printf(" \n b: From SC test, equivalent copper loss at rated load = %.1f kW, ",Psc);
+printf(" \n and the various energy losses during the 24 hr period are tabulated as :\n");
+
+printf(" \n _____________________________________________________________________________________");
+printf(" \n Percent Rated load \t Power loss(kW) \t Time period(hours) \t Energy loss(kWh)");
+printf(" \n _____________________________________________________________________________________");
+printf(" \n\t\t%d \t %f \t\t\t %d \t\t\t %.2f \n ",LF_1,P_loss_1,t1,energy_loss1);
+printf(" \n\t\t%d \t %f \t\t\t %d \t\t\t %.2f \n ",LF_2,P_loss_2,t2,energy_loss2);
+printf(" \n\t\t%d \t %f \t\t\t %d \t\t\t %.2f \n ",LF_3,P_loss_3,t3,energy_loss3);
+printf(" \n\t\t%d \t %f \t\t\t %d \t\t\t %.2f \n ",LF_fl,P_loss_fl,t_fl,energy_loss_fl);
+printf(" \n\t\t%d \t %f \t\t\t %d \t\t\t %.2f \n ",LF_4,P_loss_4,t4,energy_loss4);
+printf(" \n _____________________________________________________________________________________");
+printf(" \n Total energy load losses over 24hour period (excluding 2hrs at no-load) = %.2f ",W_loss_total);
+printf(" \n _____________________________________________________________________________________\n\n");
+
+printf(" \n c: Total energy output over the 24 hour period is tabulated as : \n");
+
+printf(" \n _____________________________________________________________________________________");
+printf(" \n Percent Rated load \t PF \t kW \t Time period(hours) \t Energy delivered(kWh)");
+printf(" \n _____________________________________________________________________________________");
+printf(" \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ",LF_1,PF1,P_1,t1,Energy_1);
+printf(" \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ",LF_2,PF2,P_2,t2,Energy_2);
+printf(" \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ",LF_3,PF3,P_3,t3,Energy_3);
+printf(" \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ",LF_fl,PF1,P_fl,t_fl,Energy_fl);
+printf(" \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ",LF_4,PF4,P_4,t4,Energy_4);
+printf(" \n _____________________________________________________________________________________");
+printf(" \n Total energy required by load for 24hour period (excluding 2hrs at no-load) = %d ",W_out_total);
+printf(" \n _____________________________________________________________________________________\n\n");
+
+printf(" \n d: All-day efficiency = %.1f percent",eta);
+
diff --git a/1092/CH14/EX14.32/Example14_32.sce b/1092/CH14/EX14.32/Example14_32.sce new file mode 100755 index 000000000..4a4641079 --- /dev/null +++ b/1092/CH14/EX14.32/Example14_32.sce @@ -0,0 +1,43 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-32
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S_1 = 10 ; // VA rating of small transformer
+V = 115 ; // voltage rating of transformer in volt
+V_2_1 = 6.3 ; // voltage rating of one part of secondary winding in volt
+V_2_2 = 5.0 ; // voltage rating of other part of secondary winding in volt
+Z_2_1 = 0.2 ; // impedance of one part of secondary winding in ohm
+Z_2_2 = 0.15 ; // impedance of other part of secondary winding in ohm
+
+
+// Calculations
+// case a
+V_2 = V_2_1 + V_2_2 ; // voltage across secondary winding in volt
+I_2 = S_1 / V_2 ; // Rated secondary current in A when the LV secondaries are
+// connected in series-aiding
+
+// case b
+I_c = (V_2_1 - V_2_2) / (Z_2_1 + Z_2_2); // Circulating current when LV windings are paralled
+percent_overload = (I_c / I_2)*100 ; // percent overload produced
+
+// Display the results
+disp("Example 14-32 Solution : ");
+
+printf(" \n a: Both coils must be series-connected and used to account for the ");
+printf(" \n full VA rating of the transformer. Hence,the rated current in 5 V ");
+printf(" \n and 6.3 V winding is : \n");
+printf(" \n I_2 = %.3f A \n\n", I_2);
+
+printf(" \n b: When the windings are paralleled, the net circulating current is ");
+printf(" \n the net voltage applied across the total internal impedance of ");
+printf(" \n the windings,or :\n");
+printf(" \n I_c = %.2f A \n ",I_c);
+
+printf(" \n The percent overload is = %f percent ≃ %.f percent ",percent_overload,percent_overload);
diff --git a/1092/CH14/EX14.33/Example14_33.sce b/1092/CH14/EX14.33/Example14_33.sce new file mode 100755 index 000000000..dad16e239 --- /dev/null +++ b/1092/CH14/EX14.33/Example14_33.sce @@ -0,0 +1,62 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-33
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S = 20 ; // kVA rating of transformer
+N_1 = 230 ; // Number of primary turns
+N_2 = 20 ; // Number of secondary turns
+
+V_1 = 230 ; // Primary voltage in volt
+V_2 = 20 ; // Secondary voltage in volt
+
+// from Fig.14-31a
+// HV side SC test data
+V_sc = 4.5 ; // short circuit voltage in volt
+I_sc = 87 ; // short circuit current in A
+P_sc = 250 ; // Power measured in W
+
+// Calculations
+// case a
+V_h = V_sc ;// short circuit voltage in volt on HV side
+I_h = I_sc ;// short circuit current in A on HV side
+Z_eh = V_h /I_h ; // Equivalent immpedance reffered to the high side when coils are series connected
+
+// case b
+Z_el = Z_eh * (N_2/N_1)^2 ; //Equivalent immpedance reffered to the low side
+// when coils are series connected
+
+// case c
+I_2_rated = (S*1000)/V_2 ; // Rated secondary current when coils are series connected
+
+// case d
+I_2_sc = S / Z_el ; // Secondary current when the coils in Fig.14-31a are
+// short-circuited with rated voltage applied to the HV side
+
+percent_overload = (I_2_sc/I_2_rated)*100 ; // percent overload
+
+
+// Display the results
+disp("Example 14-33 Solution : ");
+
+printf(" \n Slight variations in answers are due to non-approximated calculations");
+printf(" \n in scilab\n\n");
+printf(" \n a: Equivalent immpedance reffered to the high side when coils are series connected :");
+printf(" \n Z_eh = %f ohm \n ",Z_eh);
+
+printf(" \n b: Equivalent immpedance reffered to the low side when coils are series connected :");
+printf(" \n Z_el = %f ohm \n ",Z_el);
+
+printf(" \n c: Rated secondary current when coils are series connected :");
+printf(" \n I_2(rated) = %d A \n",I_2_rated);
+
+printf(" \n d: Secondary current when the coils in Fig.14-31a are short-circuited :");
+printf(" \n with rated voltage applied to the HV side :");
+printf(" \n I_2(sc) = %d A \n",I_2_sc);
+printf(" \n The percent overload is = %d percent",percent_overload);
diff --git a/1092/CH14/EX14.34/Example14_34.sce b/1092/CH14/EX14.34/Example14_34.sce new file mode 100755 index 000000000..40bdbd646 --- /dev/null +++ b/1092/CH14/EX14.34/Example14_34.sce @@ -0,0 +1,59 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-34
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+I_L = 100 ; // Load current in A
+cos_theta = 0.7 ; // power factor lagging
+
+// Y-Δ distribution transformer
+S = 60 ; // kVA rating of transformer
+V_1 = 2300 ; // primary voltage in volt
+V_2 = 230 ; // secondary voltage in volt
+
+// Calculations
+// case a
+V_L = 230 ; // voltage across load in volt
+P_T = (sqrt(3)*V_L*I_L*cos_theta)/1000 ; // power consumed by the plant in kW
+kVA_T = P_T/cos_theta ; // apparent power in kVA
+
+// case b
+kVA = S ; // kVA rating of transformer
+V_p = V_2 ; // phase voltage in volt (delta- connection on load side)
+I_P2_rated = (kVA*1000)/(3*V_p); // Rated secondary phase current in A
+I_L2_rated = sqrt(3)*I_P2_rated ; // Rated secondary line current in A
+
+// case c
+//percent load on each transformer = (load current per line) / (rated current per line)
+percent_load = I_L / I_L2_rated * 100 ;
+
+// case d
+// subscript d for V_L indicates case d ,V_L
+V_L_d = 2300 ;
+I_P1 = (kVA_T*1000)/(sqrt(3)*V_L_d); // primary phase current in A
+I_L1 = I_P1 ; // primary line current in A(Y-connection)
+
+// case e
+kVA_transformer = kVA / 3 ; // kVA rating of each transformer
+
+// Display the results
+disp("Example 14-34 Solution : ");
+
+printf(" \n a: power consumed by the plant :\n P_T = %.1f kW \n ",P_T);
+printf(" \n apparent power :\n kVA_T = %.1f kVA \n",kVA_T);
+
+printf(" \n b: Rated secondary phase current :\n I_P2(rated) = %f A ≃ %.f A \n",I_P2_rated,I_P2_rated);
+printf(" \n Rated secondary line current :\n I_L2(rated) = %f A ≃ %.1f A \n",I_L2_rated,I_L2_rated);
+
+printf(" \n c: percent load on each transformer = %.1f percent \n ",percent_load);
+
+printf(" \n d: primary phase current :\n I_P1 = %.f A \n",I_P1);
+printf(" \n primary line current :\n I_L1 = %.f A \n",I_L1);
+
+printf(" \n e: kVA rating of each transformer = %d kVA",kVA_transformer);
diff --git a/1092/CH14/EX14.35/Example14_35.sce b/1092/CH14/EX14.35/Example14_35.sce new file mode 100755 index 000000000..fabc2c776 --- /dev/null +++ b/1092/CH14/EX14.35/Example14_35.sce @@ -0,0 +1,61 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-35
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+I_L = 100 ; // Load current in A
+cos_theta = 0.7 ; // power factor lagging
+
+// Δ-Δ distribution transformer
+S = 60 ; // kVA rating of transformer
+V_1 = 2300 ; // primary voltage in volt
+V_2 = 230 ; // secondary voltage in volt
+
+// Calculations
+// case a
+V_L = 230 ; // voltage across load in volt
+P_T = (sqrt(3)*V_L*I_L*cos_theta)/1000 ; // power consumed by the plant in kW
+kVA_T = P_T/cos_theta ; // apparent power in kVA
+
+// case b
+kVA = S ; // kVA rating of transformer
+V_p = V_2 ; // phase voltage in volt
+I_P2_rated = (kVA*1000)/(3*V_p); // Rated secondary phase current in A
+I_L2_rated = sqrt(3)*I_P2_rated ; // Rated secondary line current in A
+
+// case c
+//percent load on each transformer = (load current per line) / (rated current per line)
+percent_load = I_L / I_L2_rated * 100 ;
+
+// case d
+// subscript d for V_L indicates case d ,V_L
+V_L_d = 2300 ;
+I_P1 = (kVA_T*1000)/(sqrt(3)*V_L_d); // primary phase current in A
+I_L1 = sqrt(3)*I_P1 ; // primary line current in A
+
+// case e
+kVA_transformer = kVA / 3 ; // kVA rating of each transformer
+
+// Display the results
+disp("Example 14-35 Solution : ");
+
+printf(" \n a: power consumed by the plant :\n P_T = %.1f kW \n ",P_T);
+printf(" \n apparent power :\n kVA_T = %.1f kVA \n",kVA_T);
+
+printf(" \n b: Rated secondary phase current :\n I_P2(rated) = %f A ≃ %.f A \n",I_P2_rated,I_P2_rated);
+printf(" \n Rated secondary line current :\n I_L2(rated) = %f A ≃ %.1f A \n",I_L2_rated,I_L2_rated);
+
+printf(" \n c: percent load on each transformer = %.1f percent \n ",percent_load);
+
+printf(" \n d: primary phase current :\n I_P1 = %.f A \n",I_P1);
+printf(" \n primary line current :\n I_L1 = %f A ≃ %.1f A \n",I_L1,I_L1);
+printf(" \n The primary line current drawn by a Δ-Δ bank is √3 times the ");
+printf(" \n line current drawn by a Y-Δ bank.\n");
+
+printf(" \n e: kVA rating of each transformer = %d kVA",kVA_transformer);
diff --git a/1092/CH14/EX14.36/Example14_36.sce b/1092/CH14/EX14.36/Example14_36.sce new file mode 100755 index 000000000..ae495d94e --- /dev/null +++ b/1092/CH14/EX14.36/Example14_36.sce @@ -0,0 +1,109 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-36
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase,3-wire Δ-connected transformer shown in Fig.14-42
+V_L = 33 ; // line voltage in kV
+
+f = 60 ;// frequency in Hz
+
+// power factor
+PF1 = 1; // unity power factor for I_AB
+PF2 = 0.7; // 0.7 lagging power factor for I_BC
+PF3 = 0.9; // 0.9 lagging power factor for I_CA
+
+// Calculations
+V_AB = V_L*exp(%i*(0)*(%pi/180)) ; // line voltage in kV taken as reference voltage
+
+V_BC = V_L*exp(%i*(-120)*(%pi/180)) ; // line voltage in kV
+V_BC_m = abs(V_BC);//V_BC_m=magnitude of V_BC in kV
+V_BC_a = atan(imag(V_BC) /real(V_BC))*180/%pi - 180 ;//V_BC_a=phase angle of V_BC in degrees
+// 180 is subtracted from I_BC_a to make it similar to textbook angle
+
+V_CA = V_L*exp(%i*(-240)*(%pi/180)) ; // line voltage in kV
+V_CA_m = abs(V_CA);//V_CA_m=magnitude of V_CA in kV
+V_CA_a = atan(imag(V_CA) /real(V_CA))*180/%pi - 180 ;//V_CA_a=phase angle of V_CA in degrees
+// 180 is subtracted from I_BC_a to make it similar to textbook angle
+
+theta_1 = acosd(PF1); // PF1 angle
+theta_2 = acosd(PF2); // PF2 angle
+theta_3 = acosd(PF3); // PF3 angle
+
+
+I_AB = 10*exp(%i*(theta_1)*(%pi/180)) ; // I_AB current in kA
+I_AB_m = abs(I_AB);//I_AB_m=magnitude of I_AB in kA
+I_AB_a = atan(imag(I_AB) /real(I_AB))*180/%pi;//I_AB_a=phase angle of I_AB in degrees
+
+I_BC = 15*exp(%i*(-120 - theta_2)*(%pi/180)) ; // I_BC current in kA
+I_BC_m = abs(I_BC);//I_BC_m=magnitude of I_BC in kA
+I_BC_a = atan(imag(I_BC) /real(I_BC))*180/%pi - 180;//I_BC_a=phase angle of I_BC in degrees
+// 180 is subtracted from I_BC_a to make it similar to textbook angle
+
+I_CA = 12*exp(%i*(-240 + theta_3)*(%pi/180)) ; // I_CA current in kA
+I_CA_m = abs(I_CA);//I_CA_m=magnitude of I_CA in kA
+I_CA_a = 180 + atan(imag(I_CA) /real(I_CA))*180/%pi;//I_CA_a=phase angle of I_CA in degrees
+// 180 is added to I_BC_a to make it similar to textbook angle
+
+// case a
+I_AC = -I_CA ;
+I_A = I_AB + I_AC ; // phase current in kA
+I_A_m = abs(I_A);//I_A_m=magnitude of I_A in kA
+I_A_a = atan(imag(I_A) /real(I_A))*180/%pi;//I_A_a=phase angle of I_A in degrees
+
+// case b
+I_BA = -I_AB ;
+I_B = I_BC + I_BA ; // phase current in kA
+I_B_m = abs(I_B);//I_B_m=magnitude of I_B in kA
+I_B_a = atan(imag(I_B) /real(I_B))*180/%pi;//I_B_a=phase angle of I_B in degrees
+
+// case c
+I_CB = -I_BC ;
+I_C = I_CA + I_CB ; // phase current in kA
+I_C_m = abs(I_C);//I_C_m=magnitude of I_C in kA
+I_C_a = atan(imag(I_C) /real(I_C))*180/%pi;//I_C_a=phase angle of I_C in degrees
+
+// case d
+phasor_sum = I_A + I_B + I_C ;
+
+
+// Display the results
+disp("Example 14-36 Solution : ");
+
+printf(" \n We must first write each of the phase currents in polar form. ");
+printf(" \n Since reference voltage,V_AB is assumed as 33 <0 kV, we may write\n");
+
+printf(" \n I_AB = %d <%d kA (unity PF),\n",I_AB_m,I_AB_a);
+printf(" \n But I_BC lags V_BC, which is %.f <%d kV",V_BC_m,V_BC_a);
+printf(" \n by θ = acosd(%.1f) = -%.2f lag, and consequently",PF2,theta_2);
+printf(" \n I_BC = %.f <%.2f kA \n",I_BC_m,I_BC_a);
+
+printf(" \n Similarly,I_CA leads V_CA = %.f <%.f kV",V_CA_m,V_CA_a);
+printf(" \n by θ = acosd(%.1f) = %.2f lead, and consequently",PF3,theta_3);
+printf(" \n I_CA = %d <%.2f kA \n",I_CA_m,I_CA_a);
+
+printf(" \n Writing three phase currents in comples form yields.\n");
+printf(" \n I_AB in kA = ");disp(I_AB);
+printf(" \n I_BC in kA = ");disp(I_BC);
+printf(" \n I_CA in kA = ");disp(I_CA);
+
+printf(" \n From conventional three phase theory for unbalanced Δ-connected loads");
+printf(" \n and from Fig.14-42, we have\n");
+
+printf(" \n a: I_A in kA = ");disp(I_A);
+printf(" \n I_A = %.2f <%.2f kA \n",I_A_m,I_A_a);
+
+printf(" \n b: I_B in kA = ");disp(I_B);
+printf(" \n I_B = %.2f <%.2f kA \n",I_B_m,I_B_a);
+
+printf(" \n c: I_C in kA = ");disp(I_C);
+printf(" \n I_C = %.2f <%.2f kA \n",I_C_m,I_C_a);
+
+printf(" \n d: Phasor sum of the line currents :");
+printf(" \n ΣI_L in kA = ");disp(phasor_sum);
diff --git a/1092/CH14/EX14.37/Example14_37.sce b/1092/CH14/EX14.37/Example14_37.sce new file mode 100755 index 000000000..177bbc969 --- /dev/null +++ b/1092/CH14/EX14.37/Example14_37.sce @@ -0,0 +1,56 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-37
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Δ-Δ transformers in Ex.35
+kVA_1 = 20 ; // kVA rating of transformer 1
+kVA_2 = 20 ; // kVA rating of transformer 2
+kVA_3 = 20 ; // kVA rating of transformer 3
+
+V_1 = 2300 ; // Primary voltage in volt
+V_2 = 230 ; // Secondary voltage in volt
+
+kVA = 40 ; // kVA supplied by the bank
+PF = 0.7 ; // lagging power factor at which bank supplies kVA
+
+// one defective transformer is removed
+
+// Calculations
+// case a
+kVA_transformer = kVA / sqrt(3); // kVA load carried by each transformer
+
+// case b
+percent_ratedload_Tr = kVA_transformer / kVA_1 * 100 ; // percent load carried by each transformer
+
+// case c
+kVA_V_V = sqrt(3)*kVA_1 ; // Total kVA rating of the transformer bank in V-V
+
+// case d
+ratio_banks = kVA_V_V / (kVA_1 + kVA_2 + kVA_3) * 100; // ratio of V-V bank to Δ-Δ bank Tr ratings
+
+// case e
+kVA_Tr = kVA / 3 ;
+percent_increase_load = kVA_transformer / kVA_Tr * 100 ; // percent increase in load on each transformer when one Tr is removed
+
+
+// Display the results
+disp("Example 14-37 Solution : ");
+
+printf(" \n a: kVA load carried by each transformer = %.1f kVA/transformer\n",kVA_transformer);
+
+printf(" \n b: percent rated load carried by each transformer = %.1f percent \n",percent_ratedload_Tr);
+
+printf(" \n c: Total kVA rating of the transformer bank in V-V = %.2f kVA \n",kVA_V_V);
+
+printf(" \n d: ratio of V-V bank to Δ-Δ bank Tr ratings = %.1f percent \n",ratio_banks);
+
+printf(" \n e: kVA load carried by each transformer(V-V) = %.2f kVA/transformer\n",kVA_Tr);
+printf(" \n percent increase in load on each transformer when one Tr is removed :");
+printf(" \n = %.1f percent",percent_increase_load);
diff --git a/1092/CH14/EX14.38/Example14_38.sce b/1092/CH14/EX14.38/Example14_38.sce new file mode 100755 index 000000000..c7cf1c759 --- /dev/null +++ b/1092/CH14/EX14.38/Example14_38.sce @@ -0,0 +1,46 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-38
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase SCIM
+V = 440 ; // rated voltage in volt of SCIM
+hp = 100 ; // rated power in hp of SCIM
+PF = 0.8 ; // power factor
+V_1 = 155 ; // primary voltage in volt of Tr
+V_2 = 110 ; // secondary voltage in volt of Tr
+
+V_a = 110 ; // armature voltage in volt
+V_L = 440 ; // Load voltage in volt
+eta = .98 ; // efficiency of the Tr.
+
+// Calculations
+// case a
+// referring to appendix A-3,Table 430-150 footnotes
+I_L = 124*1.25 ; // Motor line current in A
+
+// case b
+alpha = V_a/V_L ; // Transformation ratio
+
+// case c
+I_a = (sqrt(3)/2)*( I_L / (alpha*eta) ); // Current in the primary of the scott transformers
+
+// case d
+kVA = (V_a*I_a)/((sqrt(3)/2)*1000); // kVA rating of the main and teaser transformers
+
+// Display the results
+disp("Example 14-38 Solution : ");
+
+printf(" \n a: Motor line current :\n I_L = %d A \n ",I_L);
+
+printf(" \n b: Transformation ratio :\n alpha = N_1/N_2 = V_a/V_L = %.2f \n",alpha);
+
+printf(" \n c: Current in the primary of the scott transformers :\n I_a = %.f A \n",I_a);
+
+printf(" \n d: kVA rating of the main and teaser transformers :\n kVA = %.1f kVA",kVA);
diff --git a/1092/CH14/EX14.39/Example39.sce b/1092/CH14/EX14.39/Example39.sce new file mode 100755 index 000000000..2c63e2bc6 --- /dev/null +++ b/1092/CH14/EX14.39/Example39.sce @@ -0,0 +1,51 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-39
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+I_L = 1 ; // Load current in kA
+V_m = 750 ; // Peak voltage in kV
+
+// Calculations
+// case a
+V_L = (V_m)/sqrt(2); // Max. allowable Vrms in kV that may be applied to the lines using ac
+
+// case b
+S_T_ac = sqrt(3)*V_L*I_L ; // Total 3-phase apparent power in MVA
+
+// case c
+I_rms = I_L ; // rms value of load current in kA
+I_dc =I_rms*sqrt(2); // Max.allowable current in kA that can be delivered by dc transmission
+
+// case d
+V_dc = V_m ; // dc voltage in kV
+S_T_dc = V_dc*I_dc ; // Total dc apparent power delivered by two lines in MVA
+
+// case e
+S_ac_line = S_T_ac / 3 ; // Power per ac line
+
+// case f
+S_dc_line = S_T_dc / 2 ; // Power per dc line
+
+// Display the results
+disp("Example 14-39 Solution : ");
+
+printf(" \n :a Max. allowable Vrms in kV that may be applied to the lines using ac :");
+printf(" \n V_L = %.1f kV \n ",V_L);
+
+printf(" \n :b Total 3-phase apparent power :\n S_T = %.1f MVA \n",S_T_ac);
+
+printf(" \n :c Max.allowable current in kA that can be delivered by dc transmission :");
+printf(" \n I_dc = %.3f kA \n ",I_dc);
+
+printf(" \n :d Total dc apparent power delivered by two lines :\n S_T = %.1f MVA\n",S_T_dc);
+
+printf(" \n :e Power per ac line :\n S/ac line = %.1f MVA/line \n",S_ac_line);
+
+printf(" \n :f Power per dc line :\n S/dc line = %.1f MVA/line \n",S_dc_line);
diff --git a/1092/CH14/EX14.4/Example14_4.sce b/1092/CH14/EX14.4/Example14_4.sce new file mode 100755 index 000000000..576eb39d3 --- /dev/null +++ b/1092/CH14/EX14.4/Example14_4.sce @@ -0,0 +1,74 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+N_1 = 600 ; // Number of primary turns
+N_2 = 150 ; // Some number of secondary turns
+N_3 = 300 ; // Some number of secondary turns
+Z_2 = 30 ; // Resistive load in ohm across N_2
+Z_3 = 15 ; // Resistive load in ohm across N_3
+R_2 = 30 ;
+R_3 = 15 ;
+V_p = 16 ; // Primary applied voltage in volt
+cos_theta = 1 ; // unity PF
+
+// Calculations
+// case a
+Z_2_prime = Z_2 * (N_1/N_2)^2 ; // Impedance reflected to the primary by load Z_2 in ohm
+
+// case b
+Z_3_prime = Z_3 * (N_1/N_3)^2 ; // Impedance reflected to the primary by load Z_3 in ohm
+
+// case c
+// Total impedance reflected to the primary in ohm
+Z_1 = (Z_2_prime * Z_3_prime) / (Z_2_prime + Z_3_prime) ;
+
+// case d
+I_1 = V_p / Z_1 ; // Total current drawn from the supply in A
+
+// case e
+P_t = V_p * I_1 * cos_theta ; // Total power in W drwan from the supply at unity PF
+
+// case f
+V_2 = V_p * (N_2/N_1) ; // Voltage across Z_2 in volt
+P_2 = (V_2)^2 / R_2 ; // Power dissipated in load Z_2 in W
+
+// case g
+V_3 = V_p * (N_3/N_1) ; // Voltage across Z_3 in volt
+P_3 = (V_3)^2 / R_3 ; // Power dissipated in load Z_3 in W
+
+// case h
+P_total = P_2 + P_3 ; // Total power dissipated in both loads in W
+
+// Display the results
+disp("Example 14-4 Solution : ");
+
+printf(" \n a: Impedance reflected to the primary by load Z_2 : ");
+printf(" \n Z_2 = %d ohm \n ",Z_2_prime );
+
+printf(" \n b: Impedance reflected to the primary by load Z_3 : ");
+printf(" \n Z_3 = %d ohm \n ",Z_3_prime );
+
+printf(" \n c: Total impedance reflected to the primary : ");
+printf(" \n Z_1 = %.1f ohm \n ",Z_1 );
+
+printf(" \n d: Total current drawn from the supply : ");
+printf(" \n I_1 = %.1f A \n ",I_1 );
+
+printf(" \n e: Total power drawn from the supply at unity PF : ");
+printf(" \n P_t = %.1f W \n ",P_t );
+
+printf(" \n f: Voltage across Z_2 in volt :\n V_2 = %d V \n ",V_2 );
+printf(" \n Power dissipated in load Z_2 :\n P_2 = %.2f W \n",P_2 );
+
+printf(" \n g: Voltage across Z_3 in volt :\n V_3 = %d V \n ",V_3 );
+printf(" \n Power dissipated in load Z_3 :\n P_3 = %f W \n",P_3 );
+
+printf(" \n h: Total power dissipated in both loads :\n P_t = %.1f W",P_total);
diff --git a/1092/CH14/EX14.5/Example14_5.sce b/1092/CH14/EX14.5/Example14_5.sce new file mode 100755 index 000000000..4141a8c0f --- /dev/null +++ b/1092/CH14/EX14.5/Example14_5.sce @@ -0,0 +1,50 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 100 ; // Power rating of the single channel power amplifier in W
+Z_p = 3200 ; // Output impedance in ohm of the single channel power amplifier
+N_p = 1500 ; // Number of primary turns in a tapped impedance-matching transformer
+Z_L1 = 8 ; // Amplifier output in ohm using a tapped impedance-matching transformer
+Z_L2 = 4 ; // Amplifier output in ohm using a tapped impedance-matching transformer
+
+// Calculations
+// case a
+alpha = sqrt(Z_p/Z_L1) ; // Transformation ratio
+N_2 = N_p / alpha ; // Total number of secondary turns to match 8 ohm speaker
+
+// case b
+// subscript b for alpha indicates case b
+alpha_b = sqrt(Z_p/Z_L2) ; // Transformation ratio
+N_1 = N_p / alpha_b ; // Number of primary turns to match 4 ohm speaker
+
+// case c
+turns_difference = N_2 - N_1 ; // Difference in secondary and primary turns
+// subscript c for alpha indicates case c
+alpha_c = (1500/22); // Transformation ratio
+Z_L = Z_p / (alpha_c)^2 ; // Impedance that must be connected between 4 ohm and
+// 8 ohm terminals to reflect a primary impedance of 3.2 kilo-ohm
+
+// Display the results
+disp("Example 14-5 Solution : ");
+
+printf(" \n a: Transformation ratio : \n α = %d \n ",alpha );
+printf(" \n Total number of secondary turns to match 8 ohm speaker : ");
+printf(" \n N_2 = %d t \n ",N_2 );
+
+printf(" \n b: Transformation ratio : \n α = %.3f \n ",alpha_b );
+printf(" \n Number of primary turns to match 4 ohm speaker : ");
+printf(" \n N_1 = %d t \n ",N_1 );
+
+printf(" \n c: Difference in secondary and primary turns : ");
+printf(" \n N_2 - N_1 = %.f t \n ",turns_difference );
+printf(" \n Impedance that must be connected between 4 ohm and 8 ohm ");
+printf(" \n terminals to reflect a primary impedance of 3.2 kilo-ohm : ");
+printf(" \n Z_L = %.2f ohm ",Z_L );
diff --git a/1092/CH14/EX14.6/Example14_6.sce b/1092/CH14/EX14.6/Example14_6.sce new file mode 100755 index 000000000..aaf4e2f09 --- /dev/null +++ b/1092/CH14/EX14.6/Example14_6.sce @@ -0,0 +1,35 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 100 ; // Power rating of the single channel power amplifier in W
+Z_L1 = 8 ; // Amplifier output in ohm using a tapped impedance-matching transformer
+Z_L2 = 4 ; // Amplifier output in ohm using a tapped impedance-matching transformer
+P_servo = 10 ; // Power rating of the servo motor in W
+Z_servo = 0.7 ; // Impedance of the servo motor in ohm
+
+// Calculations
+root_Z_AB = sqrt(8) - sqrt(4);
+Z_AB = (root_Z_AB)^2;
+
+// Display the resulta
+disp("Example 14-6 Solution : ");
+
+printf(" \n Z_p = %dΩ*(N_p/N_1)^2 = %dΩ*(N_p/N_2)^2\n",Z_L2,Z_L1);
+printf(" \n = Z_AB * (N_p/(N_2 - N_1)^2 ) \n");
+printf(" \n Dividing each of the three numerators by N_p^2 and taking the ");
+printf(" \n square root of each term, we have\n");
+
+printf(" \n √(Z_AB)/(N_2 - N_1) = √(4)/N_1 = √(8)/N_2 \n");
+printf(" \n √(Z_AB)/(N_2 - N_1) = √(4)/N_1 - √(8)/N_2 \n");
+printf(" \n yielding, √(Z_AB) = √(8) - √(4) = %f \n",root_Z_AB);
+printf(" \n which Z_AB = (%f)^2 = %.2f Ω \n",root_Z_AB,Z_AB);
+
+
diff --git a/1092/CH14/EX14.7/Example14_7.sce b/1092/CH14/EX14.7/Example14_7.sce new file mode 100755 index 000000000..ba6a106ad --- /dev/null +++ b/1092/CH14/EX14.7/Example14_7.sce @@ -0,0 +1,79 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 10 * exp(%i * 0 * (%pi/180)); // Supply voltage of the source 10<0 V
+R_s = 1000 ; // Resistance of the source in ohm
+R_L = 10 ; // Load resistance in ohm
+Z_L = R_L ; // Load resistance in ohm
+
+// Calculations
+// case a
+alpha = sqrt(R_s/R_L) ; // Transformation ratio of the matching transformer for MPT
+
+// case b
+V_1 = V / 2 ; // Terminal voltage in volt of the source at MPT
+
+// case c
+V_2 = V_1 / alpha ; // Terminal voltage in volt across the load at MPT
+
+// case d
+I_2 = V_2 / Z_L ; // Secondary load current in A (method 1)
+I2 = V / (2*alpha*R_L) ; // Secondary load current in A (method 2)
+
+// case e
+I_1 = I_2 / alpha ; // Primary load current drawn from the source in A (method 1)
+I1 = V / (2*R_s) ; // Primary load current drawn from the source in A (method 2)
+
+// case f
+P_L = (I_2)^2 * R_L ; // Maximum power dissipated in the load in W
+
+// case g
+P_s = (I_1)^2 * R_s ; // Power dissipated internally within the source in W
+
+// case h
+P_T1 = V * I_1 * cosd(0) ; // Total power supplied by the source in W(method 1)
+
+P_T2 = P_L + P_s ; // Total power supplied by the source in W(method 2)
+
+// case i
+P_T = P_T1 ;
+eta = P_L / P_T * 100 ; // Power transfer efficiency in percent
+
+// Display the results
+disp("Example 14-7 Solution : ");
+
+printf(" \n a: Transformation ratio of the matching transformer for MPT : ");
+printf(" \n α = %d \n ",alpha );
+
+printf(" \n b: Terminal voltage of the source at MPT :\n V_1 = %d V \n",V_1);
+
+printf(" \n c: Terminal voltage across the load at MPT :\n V_2 = %.1f V \n",V_2);
+
+printf(" \n d: Secondary load current :");
+printf(" \n (method 1) :\n I_2 = %.2f A = %d mA \n ",I_2, 1000*I_2);
+
+printf(" \n (method 2) :\n I_2 = %.2f A = %d mA \n ",I2, 1000*I2);
+
+printf(" \n e: Primary load current drawn from the source : ");
+printf(" \n (method 1) :\n I_1 = %f A = %d mA \n ",I_1 , 1000*I_1 );
+printf(" \n (method 2) :\n I_1 = %f A = %d mA \n ",I1 , 1000*I1 );
+
+printf(" \n f: Maximum power dissipated in the load : ");
+printf(" \n P_L = %f W = %d mW \n",P_L , 1000*P_L );
+
+printf(" \n g: Power dissipated internally within the source : " );
+printf(" \n P_s = %f W = %d mW \n",P_s , 1000*P_s );
+
+printf(" \n h: Total power supplied by the source : ");
+printf(" \n (method 1) :\n P_T = %f W = %d mW \n ",P_T1, 1000*P_T1);
+printf(" \n (method 2) :\n P_T = %f W = %d mW \n ",P_T2, 1000*P_T2);
+
+printf(" \n i: Power transfer efficiency :\n η = %d percent ",eta );
diff --git a/1092/CH14/EX14.8/Example14_8.sce b/1092/CH14/EX14.8/Example14_8.sce new file mode 100755 index 000000000..80c8a8500 --- /dev/null +++ b/1092/CH14/EX14.8/Example14_8.sce @@ -0,0 +1,43 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// power transformer
+V = 20 ; // No-load voltage in volt
+R_s = 18 ; // Internal resistance of the power amplifier in ohm
+R_L = 8 ; // Load resistance in ohm(Speaker)
+
+// Calculations
+// case a
+V_L = ( R_L / (R_L + R_s) )* V; // Load voltage in volt
+P_L = (V_L)^2 / R_L ; // Power delivered in W to the speaker when connected
+// directly to the amplifier
+
+// case b
+alpha = sqrt(R_s/R_L); // Turns ratio of the transformer to maximize speaker power
+
+// case c
+V_2 = V / (2*alpha); // Secondary voltage in volt
+P_L2 = (V_2)^2 / R_L ; // Maximum power delivered in W to the speaker using matching
+// transformer of part b
+
+// Display the results
+disp("Example 14-8 Solution : ");
+
+printf(" \n a; Load voltage :\n V_L = %.2f V across the 8 Ω speaker\n ",V_L);
+printf(" \n Power delivered in W to the speaker when connected directly to the amplifier : ");
+printf(" \n P_L = %.2f W \n ", P_L );
+
+printf(" \n b: Turns ratio of the transformer to maximize speaker power : ");
+printf(" \n α = %.1f \n ", alpha );
+
+printf(" \n c: Secondary voltage :\n V_2 = %f V \n ",V_2 );
+printf(" \n Maximum power delivered in W to the speaker using matching ");
+printf(" \n transformer of part b :\n P_L = %f W ",P_L2 );
diff --git a/1092/CH14/EX14.9/Example14_9.sce b/1092/CH14/EX14.9/Example14_9.sce new file mode 100755 index 000000000..46385c495 --- /dev/null +++ b/1092/CH14/EX14.9/Example14_9.sce @@ -0,0 +1,42 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 1 ; // kVA rating of the transformer
+V_1 = 220 ; // Primary voltage in volt
+V_2 = 110 ; // Secondary voltage in volt
+f_o = 400 ; // Frequency in Hz(original frequency)
+f_f = 60 ; // Frequency in Hz for which the transformer is to be used
+
+// Calculations
+alpha = V_1 / V_2 ; // Transformation ratio
+// case a
+E_h = V_1 * (f_f / f_o); // Maximum rms voltage in volt applied to HV side
+E_1 = E_h ;
+E_l = E_1 / alpha ; // Maximum rms voltage in volt applied to HV side
+
+// case b
+V_h = V_1 ; // High voltage in volt
+I_h = kVA * 1000 / V_h ;
+Vh = E_h ;
+kVA_new = Vh * I_h ;
+
+// Display the results
+disp("Example 14-9 Solution : ");
+
+printf(" \n a: To maintain the same permissible flux density in Eqs.(14-15)");
+printf(" \n and (14-16),both voltages of the high and low sides must change ");
+printf(" \n in the same proportion as the frequency : ");
+printf(" \n E_h = %d V \n and,\n E_l = %.1f V\n",E_h , E_l );
+
+printf(" \n b: The original current rating of the transformer is unchanged since");
+printf(" \n the conductors still have the same current carrying capacity.");
+printf(" \n Thus,\n I_h = %.3f A\n and the new kVA rating is",I_h );
+printf(" \n V_h*I_h = V_1*I_1 = %d VA = %.2f kVA",kVA_new , kVA_new/1000);
diff --git a/1092/CH2/EX2.1/Example2_1.sce b/1092/CH2/EX2.1/Example2_1.sce new file mode 100755 index 000000000..bf34825d3 --- /dev/null +++ b/1092/CH2/EX2.1/Example2_1.sce @@ -0,0 +1,23 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+m = 3; // Multipicity of the armature
+P = 14; // No. of poles
+
+// Calculations
+a_lap = m * P; // No. of parallel paths in the armature for a lap winding
+a_wave = 2 * m; // No. of parallel paths in the armature for a wave winding
+
+// Display the result
+disp("Example 2-1 Solution : ");
+
+printf("\n a: a = %d paths ", a_lap);
+printf("\n b: a = %d paths ", a_wave);
diff --git a/1092/CH2/EX2.2/Example2_2.sce b/1092/CH2/EX2.2/Example2_2.sce new file mode 100755 index 000000000..1de69e410 --- /dev/null +++ b/1092/CH2/EX2.2/Example2_2.sce @@ -0,0 +1,32 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 14; // No. of poles
+phi = 4.2e6; // Flux per pole
+S = 60; // Generator speed
+coils = 420; // No. of coils
+turns_per_coil = 20;
+conductors_per_turn = 2;
+a_lap = 42; // No. of parallel paths in the armature for a lap winding
+a_wave = 6; // No. of parallel paths in the armature for a wave winding
+
+// Calculations
+Z = coils * turns_per_coil * conductors_per_turn; // No. of conductors
+E_g_lap = (( phi * Z * S * P ) / ( 60 * a_lap )) * 10 ^ -8; // Generated EMF for
+// lap winding ( Eq 1-5a)
+E_g_wave = ( phi * Z * S * P ) / ( 60 * a_wave ) * 10 ^ -8; // Generated EMF for
+// wave winding ( Eq 1-5a)
+
+// Display the result
+disp("Example 2-2 Solution : ");
+
+printf("\n a: Eg = %0.1f V ", E_g_lap);
+printf("\n b: Eg = %0.1f V ", E_g_wave);
diff --git a/1092/CH2/EX2.3/Example2_3.sce b/1092/CH2/EX2.3/Example2_3.sce new file mode 100755 index 000000000..693e167e9 --- /dev/null +++ b/1092/CH2/EX2.3/Example2_3.sce @@ -0,0 +1,31 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+slots = 72; // No. of slots
+P = 4; // No. of poles
+coils_spanned = 14; // 14 slots are spanned while winding the coils
+
+// Calculations
+Pole_span = slots / P; // Pole span
+p_not = coils_spanned / Pole_span * 180; // Span of the coil in
+// electrical degrees
+funcprot(0) ; // Use to avoid this message "Warning : redefining function: beta "
+beta = (180 - p_not);
+k_p1 = cosd( beta / 2 ); // Pitch factor using eq(2-7)
+k_p2 = sind( p_not / 2 ); // Pitch factor using eq(2-8)
+
+// Display the results
+disp("Example 2-3 Solution : ")
+printf(" \n a: Full-pitch coil span = %d slots/pole ", Pole_span );
+printf(" \n b: p = %d degrees ", p_not );
+printf(" \n c: kp = %.2f \t\t eq(2-7)", k_p1 );
+printf(" \n d: kp = %.2f \t\t eq(2-8)", k_p2 );
+
diff --git a/1092/CH2/EX2.4/Example2_4.sce b/1092/CH2/EX2.4/Example2_4.sce new file mode 100755 index 000000000..38c38eeb6 --- /dev/null +++ b/1092/CH2/EX2.4/Example2_4.sce @@ -0,0 +1,21 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+fractional_pitch = 13 / 16;
+slot =96; // No. of slots
+P = 6; // No. of poles
+
+// Calculation
+k_p = sind( ( fractional_pitch * 180 ) / 2 ); // Pitch factor
+
+// Display the result
+disp("Example 2-4 Solution : ")
+printf("\n kp = %.4f ", k_p );
diff --git a/1092/CH2/EX2.5/Example2_5.sce b/1092/CH2/EX2.5/Example2_5.sce new file mode 100755 index 000000000..fe7ee3132 --- /dev/null +++ b/1092/CH2/EX2.5/Example2_5.sce @@ -0,0 +1,26 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 12; // No. of poles
+theta = 360; // No. of mechanical degrees of rotation
+alpha_b = 180; // No. of electrical degrees for finding case b in the question
+
+// Calculations
+alpha = ( P * theta ) / 2; // No. of electrical degrees in one revolution
+n = alpha / 360; // No. of ac cycles
+theta_b = ( 2 * alpha_b ) / P; // No. of mechanical degrees of rotation
+// for finding case b in the question
+
+// Display the results
+disp("Example 2-5 Solution : ")
+printf("\n a: alpha = %d degrees", alpha);
+printf("\n n = %d cycles ", n);
+printf("\n b: theta = %d mechanical degrees ", theta_b );
diff --git a/1092/CH2/EX2.6/Example2_6.sce b/1092/CH2/EX2.6/Example2_6.sce new file mode 100755 index 000000000..5936ef83f --- /dev/null +++ b/1092/CH2/EX2.6/Example2_6.sce @@ -0,0 +1,54 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 4;// No. of poles
+phi = 3; // No. of phases
+slots_(1) = 12; // No. of slots for case 1
+slots_(2) = 24; // No. of slots for case 2
+slots_(3) = 48; // No. of slots for case 3
+slots_(4) = 84; // No. of slots for case 4
+
+// Calculations
+electrical_degrees = 180 * 4;
+i=1; // where i is case subscript .eg case1, case2, etc
+
+ while i<=4
+ alpha_(i) = electrical_degrees / slots_(i); // electrical degrees
+ // per slots for case i
+ n_(i) = slots_(i) / ( P * phi ); // No. of ac cycles for case 1
+ k_d(i) = sind( n_(i)*( alpha_(i) / 2 ) ) / ( n_(i) * sind( alpha_(i) / 2));
+ i=i+1;
+ end;
+
+// Display the results
+disp("Example 2-6 Solution : ")
+printf("\n a:");
+ i=1; // where i is case subscript .eg case1, case2, etc
+
+ while i<=4
+ printf("\n \t %d: alpha = %.2f degrees/slot", i , alpha_(i) );
+ printf("\n\t n = %d slots/pole-phase ", n_(i) );
+ printf("\n\t kd = %.3f ", k_d(i));
+ printf("\n");
+ i=i+1;
+ end;
+
+printf("\n\n\n b: ");
+printf("\n \t \t n \t alpha in degrees \t\t kd ");
+printf("\n \t __________________________________________________________" );
+i=1;
+
+ while i<=4
+ printf("\n \t \t %d \t %.2f \t\t\t\t%.3f ", n_(i) , alpha_(i) , k_d(i) );
+ i = i +1;
+ end;
+ printf("\n \t __________________________________________________________" );
+
diff --git a/1092/CH2/EX2.7/Example2_7.sce b/1092/CH2/EX2.7/Example2_7.sce new file mode 100755 index 000000000..937306a41 --- /dev/null +++ b/1092/CH2/EX2.7/Example2_7.sce @@ -0,0 +1,47 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+slots = 72; // No. of slots
+P = 6; // No. of poles
+phase =3; // three phase stator armature
+N_c = 20; // Number of turns per coil
+pitch = 5 / 6;
+phi = 4.8e+6; // flux per pole in lines
+S = 1200; // Rotor speed
+
+// Calculations
+f = ( P * S )/ 120; // Frequency of rotor
+
+E_g_percoil = 4.44 * phi * N_c * f *10 ^ -8; // Generated effective voltage
+// per coil of a full pitch coil
+
+N_p = ( slots / phase ) * N_c; // Total number of turns per phase
+
+n = slots / ( phase * P ); // No. os slots per pole per phase
+
+alpha = ( P * 180 ) / slots; // No. of electrical degrees between adjacent slots
+
+k_d = sind( n * alpha / 2 ) / ( n * sind( alpha / 2 ) ); // Distribution factor
+
+span = pitch * 180; // Span of the coil in electrical degrees
+
+k_p = sind( span / 2 ); // Pitch factor
+
+E_gp = 4.44 * phi * N_p * f * k_p * k_d * 10 ^ -8; // Total generated voltage
+// per phase considering kp and kd
+
+// Display the result
+disp("Example 2-7 Solution : ")
+printf("\n a: Eg/coil = %.2f V/coil", E_g_percoil );
+printf("\n b: Np = %d turns/phase ", N_p );
+printf("\n c: kd = %.3f ", k_d );
+printf("\n d: kp = %.3f ", k_p );
+printf("\n e: Egp = %.2f V/phase ", E_gp );
diff --git a/1092/CH2/EX2.8/Example2_8.sce b/1092/CH2/EX2.8/Example2_8.sce new file mode 100755 index 000000000..a3d4d8df1 --- /dev/null +++ b/1092/CH2/EX2.8/Example2_8.sce @@ -0,0 +1,28 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 2: Dynamo Construction and Windings
+// Example 2-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 8; // No. of poles
+S = 900; // Speed in revolutions / minute
+f_1 = 50; // Frequency of generated voltage from generator 1
+f_2 = 25; // Frequency of generated voltage from generator 2
+
+// Calculations
+f = ( P * S ) / 120; // Frequency of the generated voltage
+S_1 = ( 120 * f_1 ) / P; // Speed of generator(rpm) 1 to generate 50 Hz voltage
+S_2 = ( 120 * f_2 ) / P; // Speed of generator(rpm) 2 to generate 25 Hz voltage
+omega_1 = ( 4 * %pi * f_1 ) / P; // Speed of generator 1 in rad/s
+omega_2 = ( 4 * %pi * f_2 ) / P; // Speed of generator 2 in rad/s
+
+// Display the result
+disp("Example 2-8 Solution : ")
+printf("\n a: f = %d Hz ", f );
+printf("\n b: S1 = %d rpm \n S2 = %d rpm ", S_1, S_2 );
+printf("\n c: omega1 = %f rad/s \n omega2 = %f rad/s", omega_1, omega_2 );
diff --git a/1092/CH3/EX3.1/Example3_1.sce b/1092/CH3/EX3.1/Example3_1.sce new file mode 100755 index 000000000..795126605 --- /dev/null +++ b/1092/CH3/EX3.1/Example3_1.sce @@ -0,0 +1,29 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 150; // Power rating of Shunt generator in kW
+V_1 = 250; // Voltage rating of Shunt generator in V
+V_a = V_1; // Voltage rating of Shunt generator in V
+R_f = 50; // Field resistance in ohm
+R_a = 0.05; // Armature resistance in ohm
+
+// Calculations
+I_1 = ( kW * 1000 ) / V_1; // Full-load line current flowing to the load in A
+I_f = V_1 / R_f; // Field current in A
+I_a = I_f + I_1; // Armature current in A
+E_g = V_a + I_a * R_a; // Full load generated voltage in V
+
+// Display the results
+disp("Example 3-1 Solution : ")
+printf("\n a: I1 = %d A ", I_1 );
+printf("\n b: If = %d A ", I_f );
+printf("\n c: Ia = %d A ", I_a );
+printf("\n d: Eg = %.2f A ", E_g );
diff --git a/1092/CH3/EX3.10/Example3_10.sce b/1092/CH3/EX3.10/Example3_10.sce new file mode 100755 index 000000000..2030d5b22 --- /dev/null +++ b/1092/CH3/EX3.10/Example3_10.sce @@ -0,0 +1,26 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW= 50; // Power rating of the dynamo
+V = 125; // Rated voltage in V
+S = 1800; // Speed of the dynamo in rpm
+I_f =20; // Exciting field current
+Max_temp_rise = 25; // Maximum Temperature rise in degree celsius
+I_l = 400; // Load Current in A
+// INSULATION CLASS A
+// COMPOUND WINDING
+
+// Display the result
+disp("Example 3-10 Solution : ")
+printf("\n a: Since the speed is reduced in half, we must reduce the kW rating in half. Consequently, the 25kW, 900 rpm dynamo has the same size. ");
+printf("\n\n b: Since we have cut the speed in half but maintained the same kW rating, the dynamo has twice the size as the original.");
+printf("\n\n c: Half the size. ");
+printf("\n\n d: Same size. ");
diff --git a/1092/CH3/EX3.2/Example3_2.sce b/1092/CH3/EX3.2/Example3_2.sce new file mode 100755 index 000000000..49686899c --- /dev/null +++ b/1092/CH3/EX3.2/Example3_2.sce @@ -0,0 +1,37 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW =100; // Power rating of the generator in kW
+V_1 = 500; // Voltage rating of hte generator in V
+R_a = 0.03; // Armature resistance in ohm
+R_f = 125; // Shunt field resistance in ohm
+R_s = 0.01; // Series field resistance in ohm
+I_d = 54; // Diverter current in A
+
+// Calculations
+I_1 = ( kW * 1000 ) / V_1; // Full-load line current flowing to the load in A
+I_f = V_1 / R_f; // Shunt Field current in A
+I_a = I_f + I_1; // Armature current in A
+I_s = I_a - I_d; // Series Field current in A
+R_d = ( I_s * R_s ) / I_d; // Diverter resistance in ohm
+E_g = V_1 + I_a * R_a + I_s * R_s; // Generated voltage at full load in V
+
+// Display the results
+disp("Example 3-2 Solution : ")
+printf("\n a: Rd = %.4f ohm ", R_d );
+printf("\n b: Eg = %.2f V ", E_g );
+
+
+
+
+
+
+
diff --git a/1092/CH3/EX3.3/Example3_3.sce b/1092/CH3/EX3.3/Example3_3.sce new file mode 100755 index 000000000..8ed1ea702 --- /dev/null +++ b/1092/CH3/EX3.3/Example3_3.sce @@ -0,0 +1,26 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_orig = 150; // Armature voltage of the generator in V
+S_orig = 1800; // Speed of the generator in rpm
+S_final_a =2000; // Increased Speed of the generator in rpm for case a
+S_final_b =1600; // Increased Speed of the generator in rpm for case b
+
+// Calculations
+E_final_a = E_orig * ( S_final_a / S_orig ); // No- load voltage of the generator
+// generator in V for case a
+E_final_b = E_orig * ( S_final_b / S_orig ); // No- load voltage of the generator
+// generator in V for case b
+
+// Display the results
+disp("Example 3-3 Solution : ")
+printf("\n a: Efinal = %.1f V ", E_final_a );
+printf("\n b: Efinal = %.1f V ", E_final_b );
diff --git a/1092/CH3/EX3.4/Example3_4.sce b/1092/CH3/EX3.4/Example3_4.sce new file mode 100755 index 000000000..5a724a0d8 --- /dev/null +++ b/1092/CH3/EX3.4/Example3_4.sce @@ -0,0 +1,33 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S_final = 1200; // Speed of th generator in rpm
+E_orig_a = 64.3; // Armature voltage of the generator in V for case a
+E_orig_b = 82.9; // Armature voltage of the generator in V for case b
+E_orig_c = 162.3; // Armature voltage of the generator in V for case c
+
+S_orig_a = 1205; // Varied Speed of the generator in rpm for case a
+S_orig_b = 1194; // Varied Speed of the generator in rpm for case b
+S_orig_c = 1202; // Varied Speed of the generator in rpm for case c
+
+// Calculations
+E_1 = E_orig_a * ( S_final / S_orig_a ); // No- load voltage of the generator
+// generator in V for case a
+E_2 = E_orig_b * ( S_final / S_orig_b ); // No- load voltage of the generator
+// generator in V for case b
+E_3 = E_orig_c * ( S_final / S_orig_c ); // No- load voltage of the generator
+// generator in V for case c
+
+// Display the results
+disp("Example 3-4 Solution : ")
+printf("\n a: E1 = %.1f V at %d rpm ", E_1, S_final );
+printf("\n b: E2 = %.1f V at %d rpm ", E_2, S_final );
+printf("\n c: E3 = %.1f V at %d rpm ", E_3, S_final );
diff --git a/1092/CH3/EX3.5/Example3_5.sce b/1092/CH3/EX3.5/Example3_5.sce new file mode 100755 index 000000000..95b05b6be --- /dev/null +++ b/1092/CH3/EX3.5/Example3_5.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V = 125; // Rated voltage of the shunt generator in V
+R_a = 0.15; // Armature resistance in ohm
+V_a = 0; // Shunt generator is loaded progressively until the terminal voltage
+// across the load is zero volt
+I_1 = 96; // Load current in A
+I_f = 4; // Field current in A
+
+// Calculations
+I_a = I_f + I_1; // Armature current in A
+E_g = V_a + I_a * R_a ; // Voltage generated in the armature in V
+
+// Display the results
+disp("Example 3-5 Solution : ")
+printf("\n Ia = %d A ", I_a );
+printf("\n Eg = %d V ", E_g );
+
diff --git a/1092/CH3/EX3.6/Example3_6.sce b/1092/CH3/EX3.6/Example3_6.sce new file mode 100755 index 000000000..b4477ae0d --- /dev/null +++ b/1092/CH3/EX3.6/Example3_6.sce @@ -0,0 +1,22 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_n1 = 135; // No load voltage of the shunt generator in V
+V_f1 = 125; // Full load voltage of the shunt generator in V
+
+// Calculation
+VR = ( V_n1 - V_f1 ) / V_f1 * 100; // Percentage voltage regulation
+
+// Display the result
+disp("Example 3-6 Solution : ")
+printf(" \n VR = %d percent ", VR );
+
+
diff --git a/1092/CH3/EX3.7/Example3_7.sce b/1092/CH3/EX3.7/Example3_7.sce new file mode 100755 index 000000000..f073fe295 --- /dev/null +++ b/1092/CH3/EX3.7/Example3_7.sce @@ -0,0 +1,23 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+VR = 0.105; // voltage regulation
+V_f1 = 250; // Full load voltage of the shunt generator in V
+
+// Calculation
+V_n1 = V_f1 + ( V_f1 * VR ); // No-load voltage of the generator in V
+
+// Display the result
+disp("Example 3-7 Solution : ")
+printf("\n Vn1 = %.1f V ", V_n1 );
+
+
+
diff --git a/1092/CH3/EX3.8/Example3_8.sce b/1092/CH3/EX3.8/Example3_8.sce new file mode 100755 index 000000000..6df7fe8a2 --- /dev/null +++ b/1092/CH3/EX3.8/Example3_8.sce @@ -0,0 +1,30 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+N_f = 1000; // Shunt field winding turns
+N_s = 4; // Series field winding turns
+I_f = 0.2; // Field current in A
+I_a = 80; // Full load armature current in A
+R_s =0.05; // Series field resistance in ohm
+
+// Calculations
+deba_I_f_N_f = I_f * N_f;
+I_s_N_s = deba_I_f_N_f; // Series field ampere-turns
+I_s =( I_s_N_s ) / N_s; // Desired current in A in the series field required to
+// produce voltage rise
+I_d = I_a - I_s; // Diverter current in A
+R_d = ( I_s * R_s ) / I_d; // Diverter resistance in ohm
+
+// Display the result
+disp("Example 3-8 Solution : ")
+printf("\n a: IsNs = %d At ", I_s_N_s );
+printf("\n b: Rd = %.4f ohm ", R_d );
+
diff --git a/1092/CH3/EX3.9/Example3_9.sce b/1092/CH3/EX3.9/Example3_9.sce new file mode 100755 index 000000000..308564a70 --- /dev/null +++ b/1092/CH3/EX3.9/Example3_9.sce @@ -0,0 +1,37 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 3: DC Dynamo Voltage Relations - DC Generators
+// Example 3-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 60; // Power rating of the generator in kW
+V = 240; //Voltage rating of the generator in V
+I_f = 3; // Increase in the field current in A
+OC_V = 275; // Over Compounded Voltage in V
+I_l = 250; // Rated load current in A
+N_f = 200; // No. of turns per pole in the shunt field winding
+N_s = 5; // No. of turns per pole in the series field winding
+R_f = 240; // Shunt field resistance in ohm
+R_s = 0.005; // Series field resistance in ohm
+
+// Calculations
+deba_I_f_N_f = I_f * N_f;
+I_s_N_s = deba_I_f_N_f; // Series field ampere-turns
+I_s =( I_s_N_s ) / N_s; // Desired current in A in the series field required to
+// produce voltage rise
+I_d = I_l - I_s; // Diverter current in A
+R_d = ( I_s * R_s ) / I_d; // Diverter resistance in ohm
+NL_MMF = ( V / R_f )* N_f; // No-load MMF
+I_f_N_f = NL_MMF;
+FL_MMF = I_f_N_f + I_s_N_s; // Full-load MMF
+
+// Display the result
+disp("Example 3-9 Solution : ")
+printf("\n a: Rd = %.5f ohm ", R_d );
+printf("\n b: No-load MMF = %d At/pole ", NL_MMF );
+printf("\n Full-load MMF = %d At/pole ", FL_MMF );
diff --git a/1092/CH4/EX4.1/Example4_1.sce b/1092/CH4/EX4.1/Example4_1.sce new file mode 100755 index 000000000..9b3f10cdd --- /dev/null +++ b/1092/CH4/EX4.1/Example4_1.sce @@ -0,0 +1,33 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+d = 0.5; // diameter of the coil in m
+l = 0.6; // axial length of the coil in m
+B = 0.4; // flux density in T
+I = 25; // current carried by the coil in A
+theta = 60; // angle between the useful force & the interpolar ref axis in deg
+
+// Calculations
+F = B * I * l; // force developed on each coil side in N
+f = F * sind(theta); // force developed at the instant the coil lies at an angle
+// of 60 w.r.t the interpolar ref axis
+r = d / 2; // radius of the coil in m
+T_c = f * r; // torque developed in N-m
+T_c1 = T_c * 0.2248 * 3.281 ; // torque developed in lb-ft by first method
+T_c2 = T_c * 0.737562 ; // torque developed in lb-ft by second method
+
+// Display the results
+disp("Example 4-1 Solution : ")
+printf("\n a : F = %d N ", F );
+printf("\n b : f = %.2f N ", f );
+printf("\n c : Tc = %.2f N-m ", T_c );
+printf("\n d : 1.3 N-m * 0.2248 lb/N * 3.281 ft/m = %.2f lb-ft ", T_c1 );
+printf("\n 1.3 N-m * 0.737562 lb.ft/N.m = %.2f lb-ft ", T_c2 );
diff --git a/1092/CH4/EX4.10/Examlpe4_10.sce b/1092/CH4/EX4.10/Examlpe4_10.sce new file mode 100755 index 000000000..1389c299a --- /dev/null +++ b/1092/CH4/EX4.10/Examlpe4_10.sce @@ -0,0 +1,45 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+I_a_1 = 38 ; // Armature current in A @ full-load as per example 4-8a
+E_c_1 = 109.4 ; // Back EMF in volt @ full-load as per example 4-8a
+S_1 = 1800 ; // Speed in rpm @ full-load as per example 4-8a
+
+I_a_2 = 19 ; // Armature current in A @ half-loadas per example 4-8a
+E_c_2 = 113.2 ; // Back EMF in volt @ half-load as per example 4-8a
+S_2 = 1863 ; // Speed in rpm @ half-load as per example 4-8a
+
+I_a_3 = 47.5 ; // Armature current in A @ 125% overload as per example 4-8b
+E_c_3 = 107.5 ; // Back EMF in volt @ 125% overload as per example 4-8b
+S_3 = 1769 ; // Speed in rpm @ 125% overload as per example 4-8b
+
+I_a_4 = 63.6 ; // Armature current in A @ overload as per example 4-9
+E_c_4 = 104.3 ; // Back EMF in volt @ overload as per example 4-9
+S_4 = 1532 ; // Speed in rpm @ overload as per example 4-9
+
+// Calculations
+P_d_1 = E_c_1 * I_a_1 ; // Armature power developed @ full-load
+
+P_d_2 = E_c_2 * I_a_2 ; // Armature power developed @ half-load
+
+P_d_3 = E_c_3 * I_a_3 ; // Armature power developed @ 125% overload
+
+P_d_4 = E_c_4 * I_a_4 ; // Armature power developed @ overload
+
+// Display the results
+disp(" Example 4-10 Solution : ");
+printf("\n Example \t Ia \t Ec \t Speed \t Pd or (Ec*Ia)");
+printf("\n _______________________________________________________________________");
+printf("\n 4-8a \t\t %d \t %.1f \t %d \t %d W at full-load ", I_a_1,E_c_1,S_1,P_d_1);
+printf("\n 4-8a \t\t %d \t %.1f \t %d \t %.1f W at half-load ",I_a_2 , E_c_2 , S_2 , P_d_2);
+printf("\n 4-8b \t\t %.1f \t %.1f \t %d \t %d W at 125 percent overload ",I_a_3,E_c_3,S_3,P_d_3);
+printf("\n 4-9 \t\t %.1f \t %.1f \t %d \t %d W at overload ",I_a_4,E_c_4,S_4,P_d_4);
+printf("\n _______________________________________________________________________");
diff --git a/1092/CH4/EX4.11/Example4_11.sce b/1092/CH4/EX4.11/Example4_11.sce new file mode 100755 index 000000000..1e98e2f5a --- /dev/null +++ b/1092/CH4/EX4.11/Example4_11.sce @@ -0,0 +1,34 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+T_a = 6.5; // Torque in dyne-centimeters
+T_b = 10.6; // Torque in gram-centimeters
+T_c = 12.2; // Torque in ounce-inches
+
+// Calculations
+T_a_Nm = T_a * 1.416e-5 * 7.0612e-3 ; // Torque T_a in N-m
+T_b_Nm = T_b * ( 1 / 72.01 ) * 7.0612e-3 ; // Torque T_b in N-m
+T_c_Nm = T_c * 7.0612e-3 ; // Torque T_c in N-m
+
+T_a_lbft = T_a * 1.416e-5 * 5.208e-3; // Torque T_a in lb-ft
+T_b_lbft = T_b * ( 1 / 72.01 ) * 5.208e-3; // Torque T_b in lb-ft
+T_c_lbft = T_c * 5.208e-3; // Torque T_c in lb-ft
+
+// Display the results
+disp("Example 4-11 Solution : ");
+printf(" \n a : T = %.1e N-m ", T_a_Nm );
+printf(" \n T = %.1e lb-ft \n ", T_a_lbft );
+
+printf(" \n b : T = %.2e N-m ", T_b_Nm );
+printf(" \n T = %.1e lb-ft \n ", T_b_lbft );
+
+printf(" \n c : T = %.3e N-m ", T_c_Nm );
+printf(" \n T = %.2e lb-ft \n ", T_c_lbft );
diff --git a/1092/CH4/EX4.12/Example4_12.sce b/1092/CH4/EX4.12/Example4_12.sce new file mode 100755 index 000000000..34d0126e0 --- /dev/null +++ b/1092/CH4/EX4.12/Example4_12.sce @@ -0,0 +1,24 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_a = 120 ; // Rated terminal voltage of dc shunt notor in volt
+R_a = 0.2 ; // Armature resistance in ohm
+BD = 2 ; // Brush drop in volt
+I_a = 75 ; // Full load armature current in A
+
+// Calculations
+I_st = ( V_a - BD ) / R_a ; // Current @ the instant of starting in A
+percentage = I_st / I_a * 100 ; // Percentage at full load
+
+// Display the results
+disp(" Example 4-12 Solution : ");
+printf(" \n Ist = %d A ( Back EMF is zero )",I_st );
+printf(" \n Percentage at full load = %d percent ", percentage );
diff --git a/1092/CH4/EX4.13/Example4_13.sce b/1092/CH4/EX4.13/Example4_13.sce new file mode 100755 index 000000000..f7df7860b --- /dev/null +++ b/1092/CH4/EX4.13/Example4_13.sce @@ -0,0 +1,43 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-13
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_a = 120 ; // Rated terminal voltage of dc shunt notor in volt
+R_a = 0.2 ; // Armature resistance in ohm
+BD = 2 ; // Brush drop in volt
+I_a = 75 ; // Full load armature current in A
+I_a_new = 1.5 * I_a ; // armature current in A at 150% rated load
+
+E_c_a = 0 ; // Back EMF at starting
+E_c_b = ( 25 / 100 ) * V_a ; // Back EMF in volt is 25% of Va @ 150% rated load
+E_c_c = ( 50 / 100 ) * V_a ; // Back EMF in volt is 50% of Va @ 150% rated load
+
+// Calculations
+R_s_a = ( V_a - E_c_a - BD ) / I_a_new - R_a ; // Ra tapping value at starting
+// in ohm
+R_s_b = ( V_a - E_c_b - BD ) / I_a_new - R_a ; // Ra tapping value @ 25% of Va
+// in ohm
+R_s_c = ( V_a - E_c_c - BD ) / I_a_new - R_a ; // Ra tapping value @ 50% of Va
+// in ohm
+E_c_d = V_a - ( I_a * R_a + BD ) ; // Back EMF @ full-load without starting resistance
+
+// Display the results
+disp(" Example 4-13 Solution : ");
+printf(" \n a: At starting, Ec is zero ");
+printf(" \n Rs = %.2f ohm \n ", R_s_a );
+
+printf(" \n b: When back EMF in volt is 25 percent of Va @ 150 percent rated load ");
+printf(" \n Rs = %.3f ohm \n ", R_s_b );
+
+printf(" \n c: When back EMF in volt is 50 percent of Va @ 150 percent rated load ");
+printf(" \n Rs = %.3f ohm \n ", R_s_c );
+
+printf(" \n d: Back EMF at full-load without starting resistance ");
+printf(" \n Ec = %d V ", E_c_d );
diff --git a/1092/CH4/EX4.14/Example4_14.sce b/1092/CH4/EX4.14/Example4_14.sce new file mode 100755 index 000000000..a877e78d7 --- /dev/null +++ b/1092/CH4/EX4.14/Example4_14.sce @@ -0,0 +1,36 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-14
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Cumulative DC compound motor acting as shunt motor
+T_orig = 160 ; // Original torque developed in lb.ft
+I_a_orig = 140 ; // Original armature current in A
+phi_f_orig = 1.6e+6 ; // Original field flux in lines
+
+// Reconnected as a cumulative DC compound motor
+T_final_a = 190 ; // Final torque developed in lb.ft (case a)
+
+// Calculations
+phi_f = phi_f_orig * ( T_final_a / T_orig ) ; // Field flux in lines
+percentage = ( phi_f / phi_f_orig ) * 100 - 100 ; // percentage increase in flux
+
+phi_f_final = 1.1 * phi_f ; // 10% increase in load causes 10% increase in flux
+I_a_b = 154 ; // Final armature current in A (case b)
+T_f = T_final_a * ( I_a_b / I_a_orig ) * ( phi_f_final / phi_f ) ;
+// Final torque developed
+
+// Display the results
+disp(" Example 4-14 Solution : ");
+printf(" \n a: phi_f = %.1e lines \n ", phi_f );
+printf(" \n Percentage of flux increase = %.1f percent \n ", percentage );
+
+printf(" \n b: The final field flux is 1.1 * 1.9 * 10 ^ 6 lines " );
+printf(" \n (due to the 10 percent increase in load).The final torque is\n");
+printf(" \n T_f = %.1f lb-ft ", T_f );
diff --git a/1092/CH4/EX4.15/Example4_15.sce b/1092/CH4/EX4.15/Example4_15.sce new file mode 100755 index 000000000..620e27c5c --- /dev/null +++ b/1092/CH4/EX4.15/Example4_15.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-15
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+I_a_orig = 25 ; // Original armature current in A
+I_a_final = 30 ; // Final armature current in A
+T_orig = 90 ; // Original torque developed in lb.ft
+phi_orig = 1.0 ; // Original flux
+phi_final = 1.1 ; // Final flux
+
+// Calculations
+T_a = T_orig * ( I_a_final / I_a_orig ) ^ 2 ; // Final torque developed if field
+// is unsaturated
+T_b = T_orig * ( I_a_final / I_a_orig ) * ( phi_final / phi_orig ) ;
+// Final torque developed when Ia rises to 30 A and flux by 10%
+
+// Display the results
+disp(" Example 4-15 Solution : " );
+printf(" \n a: T = %.1f lb-ft \n ", T_a );
+printf(" \n b: T = %.1f lb-ft ", T_b );
diff --git a/1092/CH4/EX4.16/Example4_16.sce b/1092/CH4/EX4.16/Example4_16.sce new file mode 100755 index 000000000..435e8bfa4 --- /dev/null +++ b/1092/CH4/EX4.16/Example4_16.sce @@ -0,0 +1,43 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-16
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_a = 230 ; // Rated armature voltage in volt
+P = 10 ; // Rated power in hp
+S = 1250 ; // Rated speed in rpm
+R_A = 0.25 ; // Armature resistance in ohm
+R_p = 0.25 ; // Interpolar resistance
+BD = 5 ; // Brush voltage drop in volt
+R_s = 0.15 ; // Series field resistance in ohm
+R_sh = 230 ; // Shunt field resistance in ohm
+
+// shunt connection
+I_l = 54 ; // Line current in A at rated load
+I_ol = 4 ; // No-load line current in A
+S_o = 1810 ; // No-load speed in rpm
+
+// Calculations
+R_a = R_A + R_p ; // Effective armature resistance in ohm
+I_f = V_a / R_sh ; // Field current in A ( Shunt connection )
+I_a = I_ol - I_f ; // Armature current in A
+
+E_c_o = V_a - ( I_a * R_a + BD ); // No-load BACK EMF in volt
+E_c_full_load = V_a - ( I_l * R_a + BD ); // No-load BACK EMF in volt at full-load
+
+S_r = S_o * ( E_c_full_load / E_c_o ); // Speed at rated load
+
+P_d = E_c_full_load * I_l ; // Internal power in watts
+hp = P_d / 746 ; // Internal horse power
+
+// Display the results
+disp("Example 4-16 Solution : ");
+printf(" \n a: S_r = %d rpm\n ", S_r );
+printf(" \n b: P_d = %d W ", P_d );
+printf(" \n hp = %.1f hp ", hp );
diff --git a/1092/CH4/EX4.17/Example4_17.sce b/1092/CH4/EX4.17/Example4_17.sce new file mode 100755 index 000000000..b1dd4b159 --- /dev/null +++ b/1092/CH4/EX4.17/Example4_17.sce @@ -0,0 +1,72 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-17
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_a = 230 ; // Rated armature voltage in volt
+P = 10 ; // Rated power in hp
+S = 1250 ; // Rated speed in rpm
+R_A = 0.25 ; // Armature resistance in ohm
+R_p = 0.25 ; // Interpolar resistance
+BD = 5 ; // Brush voltage drop in volt
+R_s = 0.15 ; // Series field resistance in ohm
+R_sh = 230 ; // Shunt field resistance in ohm
+phi_1 = 1 ;// Original flux per pole
+
+// Long-shunt cumulative connection
+I_l = 55 ; // Line current in A at rated load
+phi_2 = 1.25 ; // Flux increased by 25% due to long-shunt cumulative connection
+I_ol = 4 ; // No-load line current in A
+S_o = 1810 ; // No-load speed in rpm
+
+// Calculations
+R_a = R_A + R_p ; // Effective armature resistance in ohm
+I_f = V_a / R_sh ; // Field current in A in shunt winding
+I_a = I_ol - I_f ; // Armature current in A for shunt connection
+E_c_o = V_a - ( I_a * R_a + BD ); // No-load BACK EMF in volt for shunt connection
+E_c_o1 = V_a - ( I_a * R_a + I_a * R_s + BD ); // No-load BACK EMF in volt for
+// long shunt cumulative connection
+S_n1 = S_o * ( E_c_o1 / E_c_o ); // Speed at no load
+
+I_f = V_a / R_sh ; // Field current in A in shunt winding
+I_a_lsh = I_l - I_f ; // Armature current in A
+E_c_full_load = V_a - ( I_a_lsh * R_a + BD ); // No-load BACK EMF in volt at
+// full-load for long-shunt cumulative connection
+
+E_c_full_load_lsh = V_a - ( I_a_lsh * R_a + I_a_lsh * R_s + BD ); // BACK EMF in volt
+// at full-load for long-shunt cumulative motor
+
+S_r = S_o * ( E_c_full_load / E_c_o ); // Speed at rated load for shunt connection
+S_r_lsh = S_n1 * ( E_c_full_load_lsh / E_c_o1 ) * ( phi_1 / phi_2 );
+// Speed at rated load for shunt connection
+
+P_d = E_c_full_load * I_a_lsh ; // Internal power in watts
+hp = P_d / 746 ; // Internal horse power
+
+T_shunt = ( hp * 5252 ) / S_r ; // Internal torque @ full-load for shunt motor
+
+I_a1 = I_a_lsh; // Armature current for shunt motor in A
+I_a2 = I_a_lsh; // Armature current for long-shunt cumulative motor in A
+T_comp = T_shunt * ( phi_2 / phi_1 ) * ( I_a2 / I_a1); // Internal torque
+// at full-load for long-shunt cumulative motor in A
+
+Horsepower = ( E_c_full_load_lsh * I_a_lsh ) / 746 ; // Internal horsepower of
+// compound motor based on flux increase
+
+// Display the results
+disp(" Example 4-17 Solution : ");
+printf(" \n a: S_n1 = %d rpm \n", S_n1 );
+printf(" \n b: S_r = %d rpm \n", S_r_lsh );
+printf(" \n c: Internal torque of shunt motor at full-load : ");
+printf(" \n T_shunt = %.2f lb-ft ", T_shunt );
+printf(" \n T_comp = %.2f lb-ft \n", T_comp );
+printf(" \n d: Horsepower = %.1f hp \n", Horsepower );
+printf(" \n e: The internal horsepower exceeds the rated horsepower because ");
+printf(" \n the power developed in the motor must also overcome the internal");
+printf(" \n mechanical rotational losses. ");
diff --git a/1092/CH4/EX4.18/Example4_18.sce b/1092/CH4/EX4.18/Example4_18.sce new file mode 100755 index 000000000..06fa4b366 --- /dev/null +++ b/1092/CH4/EX4.18/Example4_18.sce @@ -0,0 +1,59 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-18
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 25 ; // Power rating of a series motor in hp
+V_a = 250 ; // Rated voltage in volt
+R_a = 0.1 ; // Armature ckt resistance in ohm
+BD = 3 ; // Brush voltage drop in volt
+R_s = 0.05 ; // Series field resistance in ohm
+I_a1 = 85 ; // Armature current in A (case a)
+I_a2 = 100 ; // Armature current in A (case b)
+I_a3 = 40 ; // Armature current in A (case c)
+S_1 = 600 ; // Speed in rpm at current I_a1
+R_d = 0.05 ; // Divertor resistance in ohm
+
+// Calculations
+E_c2 = V_a - I_a2 * ( R_a + R_s ) - BD ; // BACK EMF in volt for I_a2
+E_c1 = V_a - I_a1 * ( R_a + R_s ) - BD ; // BACK EMF in volt for I_a1
+
+S_2 = S_1 * ( E_c2 / E_c1 ) * ( I_a1 / I_a2 ); // Speed in rpm at current I_a2
+
+E_c3 = V_a - I_a3 * ( R_a + R_s ) - BD ; // BACK EMF in volt for I_a3
+
+S_3 = S_1 * ( E_c3 / E_c1 ) * ( I_a1 / I_a3 ); // Speed in rpm at current I_a3
+
+// When divertor is connected in parallel to R_s
+R_sd = ( R_s * R_d ) / ( R_s + R_d ); // Effective series field resistance in ohm
+
+E_c2_new = V_a - I_a2 * ( R_a + R_sd ) - BD ; // BACK EMF in volt for I_a2
+S_2_new = S_1 * ( E_c2_new / E_c1 ) * ( I_a1 / ( I_a2 / 2 ) ); // Speed in rpm
+// at current I_a2
+
+E_c3_new = V_a - I_a3 * ( R_a + R_sd ) - BD ; // BACK EMF in volt for I_a3
+S_3_new = S_1 * ( E_c3_new / E_c1 ) * ( I_a1 / ( I_a3 / 2 ) ); // Speed in rpm
+// at current I_a3
+
+// Display the results
+disp(" Example 4-18 Solution : ");
+printf(" \n a: S_2 = %d rpm \n", S_2 );
+printf(" \n b: S_3 = %d rpm \n", S_3 );
+printf(" \n c: The effect of the divertor is to reduce the series field current");
+printf(" \n (and flux) to half their previous values. ");
+printf(" \n S_2 = %d rpm ", S_2_new );
+printf(" \n S_3 = %d rpm \n", S_3_new );
+
+printf(" \n The results may be tabulated as follows : \n ");
+printf(" \n Case \t I_a in A \t S_o in rpm \t S_d in rpm ");
+printf(" \n ________________________________________________________");
+printf(" \n 1 \t %d \t %d \t ___ ", I_a1 , S_1 );
+printf(" \n 2. \t %d \t %d \t %d ", I_a2 , S_2 , S_2_new );
+printf(" \n 3. \t %d \t %d \t %d ", I_a3 , S_3 , S_3_new );
+
diff --git a/1092/CH4/EX4.19/Example4_19.sce b/1092/CH4/EX4.19/Example4_19.sce new file mode 100755 index 000000000..018da33d4 --- /dev/null +++ b/1092/CH4/EX4.19/Example4_19.sce @@ -0,0 +1,34 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-19
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// From the calculations of Ex.4-16 , Ex.4-17 , Ex.4-18 we get no-load and
+// full-load speeds as follows
+S_n1 = 1810 ; // No-load speed in rpm (Ex.4-16)
+S_f1 = 1603 ; // Full-load speed in rpm (Ex.4-16)
+
+S_n2 = 1806 ; // No-load speed in rpm (Ex.4-17)
+S_f2 = 1231 ; // Full-load speed in rpm (Ex.4-17)
+
+S_n3 = 1311 ; // No-load speed in rpm (Ex.4-18)
+S_f3 = 505 ; // Full-load speed in rpm (Ex.4-18)
+
+// Calculations
+SR_1 = ( S_n1 - S_f1 ) / S_f1 * 100 ; // Speed regulation for shunt motor
+
+SR_2 = ( S_n2 - S_f2 ) / S_f2 * 100 ; // Speed regulation for compound motor
+
+SR_3 = ( S_n3 - S_f3 ) / S_f3 * 100 ; // Speed regulation for series motor
+
+// Display the results
+disp("Example 4-19 Solution : ");
+printf(" \n a: SR(shunt) = %.1f percent \n ", SR_1 );
+printf(" \n b: SR(compound) = %.1f percent \n ", SR_2 );
+printf(" \n c: SR(series) = %.1f percent \n ", SR_3 );
diff --git a/1092/CH4/EX4.2/Example4_2.sce b/1092/CH4/EX4.2/Example4_2.sce new file mode 100755 index 000000000..3e4f82daf --- /dev/null +++ b/1092/CH4/EX4.2/Example4_2.sce @@ -0,0 +1,30 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+d = 18 ; // diameter of hte coil in inches
+l = 24 ; // axial length of the coil in inches
+B = 24000 ; // Flux density in lines/sq.inches
+I = 26 ; // Current carried by the coil in A
+theta = 60 ; // angle between the useful force & the interpolar ref axis in deg
+
+
+// Calculations
+F = ( B * I * l * 10 ^ -7 ) / 1.13 ; // force developed on each coil side in lb
+f = F * sind(theta); // force developed at the instant the coil lies at an angle
+// of 60 w.r.t the interpolar ref axis
+r = d / 2; // radius of the coil in inches
+T_c = f * ( r * 1 / 12); // torque developed in lb.ft/conductor
+
+// Display the results
+disp("Example 4-2 Solution : ")
+printf("\n a : F = %.3f lb ", F );
+printf("\n b : f = %.2f lb ", f );
+printf("\n c : Tc = %.3f lb-ft/conductor ", T_c );
diff --git a/1092/CH4/EX4.20/Example4_20.sce b/1092/CH4/EX4.20/Example4_20.sce new file mode 100755 index 000000000..1dd8dcf9f --- /dev/null +++ b/1092/CH4/EX4.20/Example4_20.sce @@ -0,0 +1,23 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-20
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+SR = 0.1 ; // Given percent speed regulation 10% of a shunt motor
+omega_f1 = 60 * %pi ; // Full-load speed in rad/s
+
+// Calculations
+omega_n1 = omega_f1 * ( 1 + SR ); // No-load speed in rad/s
+
+S = omega_n1 * ( 1 / ( 2 * %pi )) * ( 60 / 1 ); // No-load speed in rpm
+
+// Display the results
+disp("Example 4-20 Solution : ");
+printf(" \n a: omega_n1 = %.2f \n ", omega_n1);
+printf(" \n b: S = %d rpm ", S );
diff --git a/1092/CH4/EX4.21/Example4_21.sce b/1092/CH4/EX4.21/Example4_21.sce new file mode 100755 index 000000000..d77929fe9 --- /dev/null +++ b/1092/CH4/EX4.21/Example4_21.sce @@ -0,0 +1,29 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-21
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S_int = 1603 ; // Internal rated speed in rpm (Ex.4-16)
+S_ext = 1250 ; // External rated speed in rpm (Ex.4-16)
+hp_int = 14.3 ; // Internal horsepower
+hp_ext = 10 ; // External horsepower
+
+// Calculations
+T_int = ( hp_int * 5252 ) / S_int ; // Internal torque in lb-ft
+
+T_ext = ( hp_ext * 5252 ) / S_ext ; // External torque in lb-ft
+
+// Display the results
+disp("Example 4-21 Solution : ");
+printf(" \n a: T_int = %.2f lb-ft \n ", T_int );
+printf(" \n b: T_ext = %.2f lb-ft \n ", T_ext );
+printf(" \n c: Internal horsepower is developed as a result of electromagnetic");
+printf(" \n torque produced by energy conversion. Some of the mechanical energy");
+printf(" \n is used internally to overcome mechanical losses of the motor,");
+printf(" \n reducing the torque available at its shaft to perform work.");
diff --git a/1092/CH4/EX4.22/Example4_22.sce b/1092/CH4/EX4.22/Example4_22.sce new file mode 100755 index 000000000..ccc693c5b --- /dev/null +++ b/1092/CH4/EX4.22/Example4_22.sce @@ -0,0 +1,21 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-22
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 50 ; // Power rating of the servo-motor in W
+S = 3000 ; // Full-load speed of the servo-motor in rpm
+
+// Calculation
+T_lbft = ( 7.04 * P ) / S ; // Output torque in lb-ft
+T_ounceinch = T_lbft * 192 ; // Output torque in ounce-inches
+
+// Display the result
+disp(" Example 4-22 Solution : ");
+printf(" \n T = %.1f oz.in ", T_ounceinch );
diff --git a/1092/CH4/EX4.23/Example4_23.sce b/1092/CH4/EX4.23/Example4_23.sce new file mode 100755 index 000000000..1603ea7f6 --- /dev/null +++ b/1092/CH4/EX4.23/Example4_23.sce @@ -0,0 +1,29 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-23
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 50 ; // Power rating of the servo-motor in W
+S_rpm = 3000 ; // Full-load speed of the servo-motor in rpm
+
+// Calculations
+S_rad_per_sec = S_rpm * 2 * %pi / 60 ; // Full-load speed of the servo-motor
+// in rad/s
+omega = 314.2 ; // Angular frequency in rad/s
+T_Nm = P / omega ; // Output torque in Nm
+T_ounceinch = T_Nm * ( 1 / 7.0612e-3 ) ; // Output torque in oz.in
+
+// Display the results
+disp("Example 4-23 Solution : ");
+printf(" \n a: Speed in rad/s = %.1f rad/s \n ", S_rad_per_sec );
+printf(" \n b: T = %.4f N-m \n ", T_Nm );
+printf(" \n c: T = %.1f oz.in \n ", T_ounceinch );
+printf(" \n d: Both answers are the same.");
+
+
diff --git a/1092/CH4/EX4.3/Example4_3.sce b/1092/CH4/EX4.3/Example4_3.sce new file mode 100755 index 000000000..0baa27acf --- /dev/null +++ b/1092/CH4/EX4.3/Example4_3.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+Z = 700 ; // no. of conductors
+d = 24 ; // diameter of the armature of the dc motor in inches
+l = 34 ; // axial length of the coil in inches
+B = 50000 ; // Flux density in lines/sq.inches
+I = 25 ; // Current carried by the coil in A
+
+// Calculations
+F_av = ( B * I * l * 10 ^ -7 ) / 1.13 * ( 700 * 0.7 ) ; // average force
+// developed on each coil side in lb
+r = d / 2; // radius of the coil in inches
+T_av = F_av * ( r /12 ) ; // armature average torque in lb-ft
+
+// Display the results
+disp("Example 4-3 Solution : ")
+printf("\n a : Fav = %.2f lb ", F_av );
+printf("\n b : Tav = %.2f lb-ft ", T_av );
diff --git a/1092/CH4/EX4.4/Example4_4.sce b/1092/CH4/EX4.4/Example4_4.sce new file mode 100755 index 000000000..faf2ee367 --- /dev/null +++ b/1092/CH4/EX4.4/Example4_4.sce @@ -0,0 +1,30 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+slots = 120 ; // No. of armature slots
+conductors_per_slot = 6 ;
+B = 60000 ; // Flux density in lines/sq.inches
+d = 28 ; // diameter of the armature
+l = 14 ; // axial length of the coil in inches
+A = 4 ; // No. of parallel paths
+span = 0.72 ; // Pole arcs span 72% of the armature surface
+I = 133.5 ; // Armature current in A
+
+// Calculations
+Z_Ta = slots * conductors_per_slot * span ; // No. of armature conductors
+F_t = ( B * I * l )/ ( 1.13 * 10 ^ 7 * A ) * Z_Ta ; // Force developed in lb
+r = ( d / 2 ) / 12 ; // radius of the armature in feet
+T = F_t * r ; // Tital torque developed
+
+// Display the result
+disp("Example 4-4 Solution : ")
+printf(" \n T = %d lb-ft", T );
+
diff --git a/1092/CH4/EX4.5/Example4_5.sce b/1092/CH4/EX4.5/Example4_5.sce new file mode 100755 index 000000000..67443f65d --- /dev/null +++ b/1092/CH4/EX4.5/Example4_5.sce @@ -0,0 +1,28 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+slots = 120 ; // No. of armature slots
+conductors_per_slot = 6 ;
+B = 60000 ; // Flux density in lines/sq.inches
+d = 28 ; // diameter of the armature
+l = 14 ; // axial length of the coil in inches
+A = 4 ; // No. of parallel paths
+span = 0.72 ; // Pole arcs span 72% of the armature surface
+T_a = 1500 ; // total armature torque in lb-ft
+
+// Calculation
+Z = slots * conductors_per_slot ; // No. of armature conductors
+r = ( d / 2 ) / 12 ; // radius of the armature in feet
+I_a = ( T_a * A * 1.13e7 ) / ( B * l * Z * r * span ) ; //Armature current in A
+
+// Display the result
+disp("Example 4-5 Solution : ")
+printf(" \n Ia = %.1f A ", I_a );
diff --git a/1092/CH4/EX4.6/Example4_6.sce b/1092/CH4/EX4.6/Example4_6.sce new file mode 100755 index 000000000..ff1dc93c1 --- /dev/null +++ b/1092/CH4/EX4.6/Example4_6.sce @@ -0,0 +1,27 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+T_old = 150 ; // Torque developed by a motor in N-m.
+disp("Example 4-6")
+disp("Given data : ")
+printf("\n \t\t\t phi \t I_a \t T ");
+printf("\n \t\t\t ________________________");
+printf("\n Original condition \t 1 \t 1 \t 150 N-m ");
+printf("\n New condition \t\t 0.9 \t 1.5 \t ? ");
+
+// Calculation
+T_new = T_old * ( 0.9 / 1 ) * ( 1.5 / 1 ) ; // New torque produced in N-m
+
+// Display the result
+printf("\n\n Solution : ")
+printf("\n Using the ratio method, the new torque is the product ");
+printf("\n of two new ratio changes : ");
+printf("\n T = %.1f N-m ", T_new );
diff --git a/1092/CH4/EX4.7/Example4_7.sce b/1092/CH4/EX4.7/Example4_7.sce new file mode 100755 index 000000000..30ba7663f --- /dev/null +++ b/1092/CH4/EX4.7/Example4_7.sce @@ -0,0 +1,28 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_a = 0.25 ; // Armature resistance in ohm
+BD = 3 ; // Brush contact drop in volt
+V = 120 ; // Applied voltage in volt
+E_a = 110 ; // EMF in volt at a given load
+E_b = 105 ; // EMF in volt due to application of extra load
+
+// Calculations
+I_a_a = ( V - ( E_a + BD ) ) / R_a ; // Armature current for E_a
+I_a_b = ( V - ( E_b + BD ) ) / R_a ; // Armature current for E_b
+del_E = ( ( E_a - E_b ) / E_a ) * 100 ; // % change in counter EMF
+del_I = ( ( I_a_a - I_a_b ) / I_a_a ) * 100 ; // % change in armature current
+
+// Display the result
+disp("Example 4-7 Solution : ")
+printf("\n a : Ia = %d A " , I_a_a );
+printf("\n b : At increased load \n Ia = %d A " , I_a_b );
+printf("\n c : del_Ec = %.2f percent \n del_Ia = %.2f percent " , del_E , del_I);
diff --git a/1092/CH4/EX4.8/Example4_8.sce b/1092/CH4/EX4.8/Example4_8.sce new file mode 100755 index 000000000..d51074f58 --- /dev/null +++ b/1092/CH4/EX4.8/Example4_8.sce @@ -0,0 +1,39 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+V_a = 120 ; // Rated terminal voltage of the DC motor in volt
+R_a = 0.2 ; // Armature circuit resistance in ohm
+R_sh = 60 ; // Shunt field resistance in ohm
+I_l = 40 ; // Line current in A @ full load
+BD = 3 ; // Brush voltage drop in volt
+S_orig = 1800 ; // Rated full-load speed in rpm
+
+// Calculations
+I_f = V_a / R_sh ; // Field current in A
+I_a = I_l - I_f ; // Armature current @ full load
+E_c_orig = V_a - ( I_a * R_a + BD ) ; // Back EMF @ full load
+
+I_a_a = I_a / 2 ; // Armature current @ half load
+E_c_a = V_a - ( I_a_a * R_a + BD ) ; // Back EMF @ half load
+S_a = S_orig * ( E_c_a / E_c_orig ) ; // Speed @ full load
+
+I_a_b = I_a * ( 5 / 4 ) ; // Armature current @ 125% overload
+E_c_b = V_a - ( I_a_b * R_a + BD ) ; // Back EMF @ 125% overload
+S_b = S_orig * ( E_c_b / E_c_orig ) ; // Speed @ 125% overload
+
+// Display the result
+disp("Example 4-8 Solution : ");
+
+printf(" \n a : At full load ");
+printf(" \n S = %.1f rpm " , S_a );
+
+printf(" \n b : At 125 percenytoverload ");
+printf(" \n S = %.1f rpm " , S_b );
diff --git a/1092/CH4/EX4.9/Example4_9.sce b/1092/CH4/EX4.9/Example4_9.sce new file mode 100755 index 000000000..b506a0c03 --- /dev/null +++ b/1092/CH4/EX4.9/Example4_9.sce @@ -0,0 +1,41 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 4: DC Dynamo Torque Relations-DC Motors
+// Example 4-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+I_l_orig = 40; // original Line current in A
+I_l_final = 66; // Final Line current in A
+
+phi_orig = 1;
+// field flux is increased by 12% so EMF produced and terminal
+// voltage will also increase by 12%
+phi_final = 1.12;
+
+V_a = 120;
+R_sh_orig = 60; // Original Field ckt resistance in ohm
+R_sh_final = 50 ; // Decreased final field ckt resistance in ohm
+
+R_a = 0.2; // Armature resistance in ohm
+BD = 3; // Brush voltage drop in volt
+S_orig = 1800; // Rated full-load speed
+
+// Calculations
+I_f_orig = V_a / R_sh_orig ; // Original Field current in A
+I_a_orig = I_l_orig - I_f_orig ; // Original Armature current @ full load
+E_c_orig = V_a - ( I_a_orig * R_a + BD ) ; // Back EMF @ full load
+
+I_f_final = V_a / R_sh_final ; // Final field current in A
+I_a_final = I_l_final - I_f_final ; // Final Armature current in A
+E_c_final = V_a - ( I_a_final * R_a + BD ) ; // Final EMF induced
+S = S_orig * ( E_c_final / E_c_orig ) * ( phi_orig / phi_final ) ;
+// Final speed of the motor
+
+// Display the result
+disp("Example 4-9 Solution : ");
+printf(" \n S = %.1f rpm ", S );
diff --git a/1092/CH5/EX5.1/Example5_1.sce b/1092/CH5/EX5.1/Example5_1.sce new file mode 100755 index 000000000..8b6684963 --- /dev/null +++ b/1092/CH5/EX5.1/Example5_1.sce @@ -0,0 +1,30 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 5: ARMATURE REACTION AND COMMUTATION IN DYNAMOS
+// Example 5-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+conductors = 800 ; // No. of conductors
+I_a = 1000 ; // Rated armature current in A
+P = 10 ; // No. of poles
+pitch = 0.7 ; // Pole-face covers 70% of the pitch
+a = P ; // No. of parallel paths ( Simplex lap-wound )
+
+// Calculations
+// Using Eq.(5-1)
+Z = conductors / P ; // No. of armature conductors/path under each pole
+Z_a = Z * pitch ; // Active armature conductors/pole
+
+// Solving for Z_p using Z_p = Z_a / a
+Z_p = Z_a / a ; // No. of pole face conductors/pole
+
+// Display the results
+disp("Example 5-1 Solution : ");
+printf(" \n No. of pole face conductors/pole to give full armature reaction ");
+printf(" \n compensation, if the pole covers 70 percent of the pitch is : \n ");
+printf(" \n Z_p = %.1f conductors/pole ", Z_p );
diff --git a/1092/CH5/EX5.2/Example5_2.sce b/1092/CH5/EX5.2/Example5_2.sce new file mode 100755 index 000000000..3420d2b48 --- /dev/null +++ b/1092/CH5/EX5.2/Example5_2.sce @@ -0,0 +1,66 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 5: ARMATURE REACTION AND COMMUTATION IN DYNAMOS
+// Example 5-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+conductors = 800 ; // No. of conductors
+I_a = 1000 ; // Rated armature current in A
+I_l = I_a ; // load or total current entering the armature in A
+P = 10 ; // No. of poles
+pitch = 0.7 ; // Pole-face covers 70% of the pitch
+a = P ; // No. of parallel paths ( Simplex lap-wound )
+alpha = 5 ; // No. of electrical degress that the brushes are shifted
+
+// Calculations
+Z = conductors / P ; // No. of armature conductors/path under each pole
+A_Z_per_pole = ( Z * I_l ) / ( P * a ); // Cross magnetizing
+// ampere-conductors/pole
+
+At_per_pole = ( 1 / 2 ) * ( 8000 / 1 ); // Ampere-turns/pole
+
+frac_demag_At_per_pole = (2*alpha) / 180 * (At_per_pole);
+// Fraction of demagnetizing ampere-turns/pole
+
+funcprot(0); // to avoid redefining function: beta warning message
+
+beta = 180 - 2*alpha ; // cross-magnetizing electrical degrees
+
+cross_mag_At_per_pole = (beta/180)*(At_per_pole);
+// cross-magnetizing ampere-turns/pole
+
+// Display the results
+disp("Example 5-2 Solution : ");
+printf(" \n a: With the brushes on the GNA,the entire armature reaction effect");
+printf(" \n is completely cross-magnetizing. The cross-magnetizing ");
+printf(" \n ampere-conductors/pole are ");
+printf(" \n = %d ampere-conductots/pole \n", A_Z_per_pole);
+
+printf(" \n and since there are 2 conductors/turn, the cross-magnetizing ");
+printf(" \n ampere-turns/pole are \n = %d At/pole \n\n", At_per_pole );
+
+
+printf(" \n b: Let alpha = the no. of electrical degrees that the brushes are ");
+printf(" \n shifted. Then the total no. of demagnetizing electrical degrees ");
+printf(" \n are 2*alpha, while the (remaining) cross-magnetizing electrical");
+printf(" \n degrees,beta, are 180 - 2*alpha. The ratio of demagnetizing to ");
+printf(" \n cross-magnetizing ampere-turns is always 2*alpha/beta. The ");
+printf(" \n fraction of demagnetizing ampere-turns/pole is ");
+printf(" \n = %.1f At/pole \n\n",frac_demag_At_per_pole );
+printf(" \n Note: Slight calculation mistake in the textbook for case b\n")
+
+
+printf(" \n c: Since beta = 180-2*alpha = 170, the cross-magnetizing ampere-turns/pole ");
+printf(" \n are \n = %.1f At/pole ", cross_mag_At_per_pole );
+
+
+
+
+
+
+
diff --git a/1092/CH6/EX6.1/Example6_1.sce b/1092/CH6/EX6.1/Example6_1.sce new file mode 100755 index 000000000..8afa90d79 --- /dev/null +++ b/1092/CH6/EX6.1/Example6_1.sce @@ -0,0 +1,52 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 1000 ; // kVA rating of the 3-phase alternator
+V_L = 4600 ; // Rated line voltage in volt
+// 3-phase, Y-connected alternator
+R_a = 2 ; // Armature resistance in ohm per phase
+X_s = 20 ; // Synchronous armature reactance in ohm per phase
+cos_theta_a = 1 ; // Unity power factor (case a)
+cos_theta_b = 0.75 ; // 0.75 power factor lagging (case b)
+sin_theta_b = sqrt( 1 - (cos_theta_b)^2 );
+
+// Calculations
+V_P = V_L / sqrt(3) ; // Phase voltage in volt
+I_P = ( kVA * 1000 ) / ( 3*V_P ) ; // Phase current in A
+I_a = I_P ; // Armature current in A
+
+// a: At unity PF
+E_g_a = ( V_P + I_a * R_a ) + %i*(I_a*X_s);
+// Full-load generated voltage per-phase (case a)
+E_g_a_m=abs(E_g_a);//E_g_a_m=magnitude of E_g_a in volt
+E_g_a_a=atan(imag(E_g_a) /real(E_g_a))*180/%pi;//E_g_a_a=phase angle of E_g_a in degrees
+
+// b: At 0.75 PF lagging
+E_g_b = ( V_P*cos_theta_b + I_a * R_a ) + %i*( V_P*sin_theta_b + I_a*X_s );
+// Full-load generated voltage per-phase (case b )
+E_g_b_m=abs(E_g_b);//E_g_b_m=magnitude of E_g_b in volt
+E_g_b_a=atan(imag(E_g_b) /real(E_g_b))*180/%pi;//E_g_b_a=phase angle of E_g_b in degrees
+
+
+// Display the results
+disp("Example 6-1 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.", sqrt(3));
+printf("\n\n a: At unity PF, \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_a);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_a_m , E_g_a_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_a_m,E_g_a_a);
+
+printf(" \n b: At 0.75 PF lagging , \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_b);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_b_m , E_g_b_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_b_m,E_g_b_a);
diff --git a/1092/CH6/EX6.2/Example6_2.sce b/1092/CH6/EX6.2/Example6_2.sce new file mode 100755 index 000000000..fecdf70b8 --- /dev/null +++ b/1092/CH6/EX6.2/Example6_2.sce @@ -0,0 +1,53 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 1000 ; // kVA rating of the 3-phase alternator
+V_L = 4600 ; // Rated line voltage in volt
+// 3-phase, Y-connected alternator
+R_a = 2 ; // Armature resistance in ohm per phase
+X_s = 20 ; // Synchronous armature reactance in ohm per phase
+cos_theta_a = 0.75 ; // 0.75 PF leading (case a)
+cos_theta_b = 0.40 ; // 0.40 PF leading (case b)
+sin_theta_a = sqrt( 1 - (cos_theta_a)^2 ); // (case a)
+sin_theta_b = sqrt( 1 - (cos_theta_b)^2 ); // (case b)
+
+// Calculations
+V_P = V_L / sqrt(3) ; // Phase voltage in volt
+I_P = ( kVA * 1000 ) / ( 3*V_P ) ; // Phase current in A
+I_a = I_P ; // Armature current in A
+
+// a: At 0.75 PF leading
+E_g_a = ( V_P*cos_theta_a + I_a * R_a ) + %i*( V_P*sin_theta_a - I_a*X_s);
+// Full-load generated voltage per-phase (case a)
+E_g_a_m=abs(E_g_a);//E_g_a_m=magnitude of E_g_a in volt
+E_g_a_a=atan(imag(E_g_a) /real(E_g_a))*180/%pi;//E_g_a_a=phase angle of E_g_a in degrees
+
+// b: At 0.40 PF leading
+E_g_b = ( V_P*cos_theta_b + I_a * R_a ) + %i*( V_P*sin_theta_b - I_a*X_s );
+// Full-load generated voltage per-phase (case b )
+E_g_b_m=abs(E_g_b);//E_g_b_m=magnitude of E_g_b in volt
+E_g_b_a=atan(imag(E_g_b) /real(E_g_b))*180/%pi;//E_g_b_a=phase angle of E_g_b in degrees
+
+
+// Display the results
+disp("Example 6-2 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.", sqrt(3));
+printf("\n\n a: 0.75 PF leading, \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_a);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_a_m , E_g_a_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_a_m,E_g_a_a);
+
+printf(" \n b: At 0.40 PF leading , \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_b);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_b_m , E_g_b_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_b_m,E_g_b_a);
diff --git a/1092/CH6/EX6.3/Example6_3.sce b/1092/CH6/EX6.3/Example6_3.sce new file mode 100755 index 000000000..ad2f089a5 --- /dev/null +++ b/1092/CH6/EX6.3/Example6_3.sce @@ -0,0 +1,41 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// From Ex.6-1 and Ex.6-2 we have V_P and E_g values as follows
+// Note : approximated values are considered when root 3 value is taken as 1.73
+// as in textbook
+V_P = 2660 ; // Phase voltage
+E_g_a1 = 3836 ; // E_g at unity PF (Ex.6-1 case a)
+E_g_b1 = 4814 ; // E_g at 0.75 PF lagging (Ex.6-1 case b)
+
+E_g_a2 = 2364 ; // E_g at 0.75 PF leading (Ex.6-2 case a)
+E_g_b2 = 1315 ; // E_g at 0.40 PF leading (Ex.6-2 case b)
+
+// Calculations
+VR_a = ( E_g_a1 - V_P )/V_P * 100 ; // voltage regulation at unity PF (Ex.6-1 case a)
+VR_b = ( E_g_b1 - V_P )/V_P * 100 ; // voltage regulation at 0.75 PF lagging (Ex.6-1 case b)
+
+VR_c = ( E_g_a2 - V_P )/V_P * 100 ; // voltage regulation at 0.75 PF leading (Ex.6-2 case a)
+VR_d = ( E_g_b2 - V_P )/V_P * 100 ; // voltage regulation at 0.40 PF leading (Ex.6-2 case b)
+
+// Display the results
+disp("Example 6-3 Solution : ");
+printf(" \n a: At unity PF : ");
+printf(" \n VR = %.1f percent \n ", VR_a );
+
+printf(" \n b: At 0.75 PF lagging : ");
+printf(" \n VR = %.2f percent \n ", VR_b );
+
+printf(" \n c: At 0.75 PF leading : ");
+printf(" \n VR = %.2f percent \n ", VR_c );
+
+printf(" \n d: At 0.40 PF leading : ");
+printf(" \n VR = %.1f percent \n ", VR_d );
diff --git a/1092/CH6/EX6.4/Example6_4.sce b/1092/CH6/EX6.4/Example6_4.sce new file mode 100755 index 000000000..1e9a42df0 --- /dev/null +++ b/1092/CH6/EX6.4/Example6_4.sce @@ -0,0 +1,77 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta_b1 = 0.8 ; // 0.8 PF lagging (case b)
+cos_theta_b2 = 0.8 ; // 0.8 PF leading (case b)
+sin_theta_b1 = sqrt( 1 - (cos_theta_b1)^2 ); // (case b)
+sin_theta_b2 = sqrt( 1 - (cos_theta_b2)^2 ); // (case b)
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculations
+// Assuming that the alternator is Y-connected
+// case a :
+I_a_rated = (kVA*1000)/(V_L*sqrt(3)); // Rated current per phase in A
+I_a = sqrt(3)*I_a_rated ; // Rated Line current in A
+
+R_dc = V_l/(2*I_a1); // effective dc armature resistance in ohm/winding
+R_ac = R_dc * 1.5 ; // effective ac armature resistance in ohm.phase
+R_a = R_ac ; // effective ac armature resistance in ohm.phase from dc resistance test
+
+Z_p = E_gp2 / I_a ; // Synchronous impedance per phase
+X_s = sqrt( Z_p^2 - R_a^2 ); // Synchronous reactance per phase
+
+// case b :
+V_p = V_L / sqrt(3); // Phase voltage in volt (Y-connection)
+
+// At 0.8 PF lagging
+E_gp1 = ( V_p*cos_theta_b1 + I_a_rated * R_a ) + %i*( V_p*sin_theta_b1 + I_a_rated * X_s);
+E_gp1_m=abs(E_gp1);//E_gp1_m=magnitude of E_gp1 in volt
+E_gp1_a=atan(imag(E_gp1) /real(E_gp1))*180/%pi;//E_gp1_a=phase angle of E_gp1 in degrees
+V_n1 = E_gp1_m ; // No-load voltage in volt
+V_f1 = V_p ; // Full-load voltage in volt
+VR1 = ( V_n1 - V_f1 )/ V_f1 * 100; // percent voltage regulation at 0.8 PF lagging
+
+
+// At 0.8 PF leading
+E_gp2 = ( V_p*cos_theta_b2 + I_a_rated * R_a ) + %i*( V_p*sin_theta_b2 - I_a_rated*X_s);
+E_gp2_m=abs(E_gp2);//E_gp2_m=magnitude of E_gp2 in volt
+E_gp2_a=atan(imag(E_gp2) /real(E_gp2))*180/%pi;//E_gp2_a=phase angle of E_gp2 in degrees
+V_n2 = E_gp2_m ; // No-load voltage in volt
+V_f2 = V_p ; // Full-load voltage in volt
+VR2 = ( V_n2 - V_f2 )/V_f2 * 100 ; // percent voltage regulation at 0.8 PF leading
+
+// Display the results
+disp("Example 6-4 Solution : ");
+printf(" \n Assuming that the alternator is Y-connected ");
+printf(" \n a: R_dc = %.1f ohm/winding ", R_dc );
+printf(" \n R_ac = %.2f ohm/phase ", R_ac );
+printf(" \n Z_p = %.2f ohm/phase ", Z_p );
+printf(" \n X_s = %.2f ohm/phase \n", X_s );
+
+printf(" \n b: At 0.8 PF lagging ");
+printf(" \n Percent voltage regulation = %.1f percent \n", VR1 );
+
+printf(" \n At 0.8 PF leading ");
+printf(" \n Percent voltage regulation = %.1f percent ", VR2 );
diff --git a/1092/CH6/EX6.5/Example6_5.sce b/1092/CH6/EX6.5/Example6_5.sce new file mode 100755 index 000000000..2dfac51b0 --- /dev/null +++ b/1092/CH6/EX6.5/Example6_5.sce @@ -0,0 +1,88 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta_b1 = 0.8 ; // 0.8 PF lagging (case b)
+cos_theta_b2 = 0.8 ; // 0.8 PF leading (case b)
+sin_theta_b1 = sqrt( 1 - (cos_theta_b1)^2 ); // (case b)
+sin_theta_b2 = sqrt( 1 - (cos_theta_b2)^2 ); // (case b)
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculations
+// Assuming that the alternator is delta-connected
+// case a :
+I_a_rated = (kVA*1000)/(V_L*sqrt(3)); // Rated current per phase in A
+I_L = I_a_rated ; // Line current in A
+
+V_p = E_gp2 ; // Phase voltage in volt
+V_l = V_p ; // Line voltage in volt (from short circuit data)
+
+I_p = I_L / sqrt(3) ; // Phase current in A (delta connection)
+I_a = I_p ; // Rated current in A
+
+Z_s = V_l / I_p ; // Synchronous impedance per phase
+R_dc = E_gp1/(2*I_a1); // effective dc armature resistance in ohm/winding
+R_ac = R_dc * 1.5 ; // effective ac armature resistance in ohm.phase
+
+// R_eff in delta = 3 * R_eff in Y
+R_eff = 3 * R_ac ; // Effective armature resistance in ohm
+R_a = R_eff ; // effective ac armature resistance in ohm.phase from dc resistance test
+
+X_s = sqrt( Z_s^2 - R_a^2 ); // Synchronous reactance per phase
+
+V_p = V_L ; // Phase voltage in volt (delta-connection)
+
+// At 0.8 PF lagging
+E_gp1 = ( V_p*cos_theta_b1 + I_a * R_a ) + %i*( V_p*sin_theta_b1 + I_a*X_s);
+E_gp1_m=abs(E_gp1);//E_gp1_m=magnitude of E_gp1 in volt
+E_gp1_a=atan(imag(E_gp1) /real(E_gp1))*180/%pi;//E_gp1_a=phase angle of E_gp1 in degrees
+V_n1 = E_gp1_m ; // No-load voltage in volt
+V_f1 = V_p ; // Full-load voltage in volt
+VR1 = ( V_n1 - V_f1 )/ V_f1 * 100; // percent voltage regulation at 0.8 PF lagging
+
+
+// At 0.8 PF leading
+E_gp2 = ( V_p*cos_theta_b2 + I_a * R_a ) + %i*( V_p*sin_theta_b2 - I_a*X_s);
+E_gp2_m=abs(E_gp2);//E_gp2_m=magnitude of E_gp2 in volt
+E_gp2_a=atan(imag(E_gp2) /real(E_gp2))*180/%pi;//E_gp2_a=phase angle of E_gp2 in degrees
+V_n2 = E_gp2_m ; // No-load voltage in volt
+V_f2 = V_p ; // Full-load voltage in volt
+VR2 = ( V_n2 - V_f2 )/V_f2 * 100 ; // percent voltage regulation at 0.8 PF leading
+
+// Display the results
+disp("Example 6-5 Solution : ");
+printf(" \n Assuming that the alternator is delta-connected : \n ");
+printf(" \n a: I_p = %.3f A ", I_p );
+printf(" \n Z_s = %.2f ohm/phase ", Z_s );
+printf(" \n R_eff in delta = %.2f ohm/phase ", R_eff );
+printf(" \n X_s = %.1f ohm/phase \n", X_s );
+printf(" \n R_eff, reactance and impedance per phase in delta is 3 times")
+printf(" \n the value when connected in Y. \n")
+
+printf(" \n b: At 0.8 PF lagging ");
+printf(" \n Percent voltage regulation = %.1f percent \n", VR1 );
+
+printf(" \n At 0.8 PF leading ");
+printf(" \n Percent voltage regulation = %.1f percent \n", VR2 );
+printf(" \n Percentage voltage regulation remains the same both in Y and delta connection.");
diff --git a/1092/CH6/EX6.6/Example6_6.sce b/1092/CH6/EX6.6/Example6_6.sce new file mode 100755 index 000000000..74dbed591 --- /dev/null +++ b/1092/CH6/EX6.6/Example6_6.sce @@ -0,0 +1,51 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase Y-connected alternator
+E_L = 11000 ; // Line voltage generated in volt
+kVA = 165000 ; // kVA rating of the alternator
+R_p = 0.1 ; // Armature resistance in ohm/per phase
+Z_p = 1.0 ; // Synchronous reactance/phase
+Z_r = 0.8 ; // Reactor reactance/phase
+
+// Calculations
+E_p = E_L / sqrt(3); // Rated phase voltage in volt
+I_p = (kVA * 1000)/(3*E_p); // Rated current per phase in A
+
+// case a
+I_max_a = E_p / R_p ; // Maximum short-circuit current in A (case a)
+overload_a = I_max_a / I_p ; // Overload (case a)
+
+// case b
+I_steady = E_p / Z_p ; // Sustained short-circuit current in A
+overload_b = I_steady / I_p ; // Overload (case b)
+
+// case c
+Z_t = R_p + %i*Z_r ; // Total reactance per phase
+I_max_c = E_p / Z_t ; // Maximum short-circuit current in A (case b)
+I_max_c_m=abs(I_max_c);//I_max_c_m=magnitude of I_max_c in A
+I_max_c_a=atan(imag(I_max_c) /real(I_max_c))*180/%pi;//I_max_c_a=phase angle of I_max_c in degrees
+overload_c = I_max_c_m / I_p ; // Overload (case a)
+
+// Display the results
+disp("Example 6-6 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.\n", sqrt(3));
+printf(" \n a: I_max = %d A ", I_max_a );
+printf(" \n overload = %.1f * rated current \n", overload_a );
+
+printf(" \n b: I_steady = %d A ", I_steady );
+printf(" \n overload = %.2f * rated current \n", overload_b );
+
+printf(" \n c: Rectangular form :\n I_max = "); disp(I_max_c);
+printf(" \n Polar form :");
+printf(" \n I_max = %d <%.2f A ", I_max_c_m , I_max_c_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",I_max_c_m,I_max_c_a);
+printf(" \n overload = %.3f * rated current \n", overload_c );
diff --git a/1092/CH6/EX6.7/Example6_7.sce b/1092/CH6/EX6.7/Example6_7.sce new file mode 100755 index 000000000..81d6cae8d --- /dev/null +++ b/1092/CH6/EX6.7/Example6_7.sce @@ -0,0 +1,70 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta = 0.8 ; // 0.8 PF lagging
+sin_theta = sqrt( 1 - (cos_theta)^2 ); //
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculated data from Ex.6-4
+I_L = 52.5 ; // Rated line current in A
+I_a = I_L ; // Rated current per phase in A
+E_gp = 532 + %i*623 ; // Generated voltage at 0.8 PF lagging
+X_s = 4.6 ; // Synchronous reactance per phase
+V_p = 635 ; // Phase voltage in volt
+
+// Calculations
+// case a
+P_T = sqrt(3) * V_L * I_L * cos_theta ; // Total output 3-phase power
+
+// case b
+P_p_b = P_T / 3 ; // Total output 3-phase power per phase
+
+// case c
+E_gp_m=abs(E_gp);//E_gp_m=magnitude of E_gp in volt
+E_gp_a=atan(imag(E_gp) /real(E_gp))*180/%pi;//E_gp_a=phase angle of E_gp in degrees
+
+// case d
+theta = acos(0.8)*180/%pi; // phase angle for PF in degrees
+theta_plus_deba = E_gp_a ; // phase angle of E_gp in degrees
+deba = theta_plus_deba - theta ; // Torque angle in degrees
+
+// case e
+P_p_e = (E_gp_m/X_s)*V_p*sind(deba); // Approximate output power/phase (Eq.(6-10))
+
+// case f
+P_p_f = E_gp_m * I_a * cosd(theta_plus_deba); // Approximate output power/phase (Eq.(6-9))
+
+// Display the results
+disp("Example 6-7 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.\n", sqrt(3));
+printf(" \n a: P_T = %d W \n", P_T );
+printf(" \n b: P_p = %.2f W \n", P_p_b );
+printf(" \n c: E_gp = %d <%.2f V \n", E_gp_m, E_gp_a );
+printf(" \n where %d is magnitude in V and %.2f is phase angle in degrees.\n",E_gp_m,E_gp_a);
+printf(" \n d: Torque angle, deba = %.2f degrees \n", deba );
+printf(" \n e: P_p = %d W \n", P_p_e );
+printf(" \n f: P_p = %d W ", P_p_f );
+
+
diff --git a/1092/CH6/EX6.8/Example6_8.sce b/1092/CH6/EX6.8/Example6_8.sce new file mode 100755 index 000000000..c3afce4d2 --- /dev/null +++ b/1092/CH6/EX6.8/Example6_8.sce @@ -0,0 +1,68 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+S = 1200 ; // Synchronous speed in rpm
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta = 0.8 ; // 0.8 PF lagging
+sin_theta = sqrt( 1 - (cos_theta)^2 ); //
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculated data from Ex.6-4 & Ex.6-7
+I_L = 52.5 ; // Rated line current in A
+I_a = I_L ; // Rated current per phase in A
+E_gp = 532 + %i*623 ; // Generated voltage at 0.8 PF lagging
+E_g = 819 ; // E_g = magnitude of E_gp in volt
+X_s = 4.6 ; // Synchronous reactance per phase
+V_p = 635 ; // Phase voltage in volt
+deba = 12.63 ; // Torque angle in degrees
+
+// Calculations
+// case a
+T_p_a = ( 7.04 * E_g * V_p * sind(deba) ) / (S*X_s); // Output torque per phase in lb.ft
+T_3phase_a = 3 * T_p_a ; // Output torque for 3-phase in lb.ft
+
+// case b
+omega = S * 2*%pi *(1/60); // Angular frequency in rad/s
+T_p_b = ( E_g * V_p * sind(deba))/(omega*X_s); // Output torque per phase in lb.ft
+T_3phase_b = 3 * T_p_b ; // Output torque for 3-phase in lb.ft
+
+// case c
+T_p_c = T_p_a * 1.356 ; // Output torque per phase in N.m
+T_3phase_c = 3 * T_p_c ; // Output torque for 3-phase in N.m
+
+// Display the results
+disp("Example 6-8 Solution : ");
+pi = %pi;
+printf(" \n Slight variations in the answers are due to value of pi = %f ",pi);
+printf(" \n and omega = %f, which are slightly different as in the textbook.\n",omega);
+printf(" \n a: T_p = %d lb-ft ",T_p_a);
+printf(" \n T_3phase = %d lb-ft \n", T_3phase_a);
+
+printf(" \n b: T_p = %.1f N-m ",T_p_b);
+printf(" \n T_3phase = %.1f N-m \n", T_3phase_b);
+
+printf(" \n c: T_p = %.1f N-m ",T_p_c);
+printf(" \n T_3phase = %.1f N-m \n", T_3phase_c);
+printf(" \n Answers from cases b and c almost tally each other ");
diff --git a/1092/CH7/EX7.1/Example7_1.sce b/1092/CH7/EX7.1/Example7_1.sce new file mode 100755 index 000000000..1d81bfcd9 --- /dev/null +++ b/1092/CH7/EX7.1/Example7_1.sce @@ -0,0 +1,74 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_sh = 120 ; // Shunt field resistance in ohm
+R_a = 0.1 ; // Armature resistance in ohm
+V_L = 120 ; // Line voltage in volt
+E_g1 = 125 ; // Generated voltage by dynamo A
+E_g2 = 120 ; // Generated voltage by dynamo B
+E_g3 = 114 ; // Generated voltage by dynamo C
+
+// Calculations
+// case a
+// 1:
+I_gA = ( E_g1 - V_L ) / R_a ; // Current in the generating source A ( in A)
+I_f = V_L / R_sh ; // Shunt field current in A
+I_a1 = I_gA + I_f ; // Armature current in A for generator A
+I_L1 = I_gA ; // Current delivered by dynamo A to the bus in A
+
+// 2:
+I_gB = ( E_g2 - V_L ) / R_a ; // Current in the generating source B ( in A)
+I_a2 = I_gB + I_f ; // Armature current in A for generator B
+I_L2 = I_gB ; // Current delivered by dynamo B to the bus in A
+
+// 3:
+I_gC = ( V_L - E_g3 ) / R_a ; // Current in the generating source C ( in A)
+I_a3 = I_gC ; // Armature current in A for generator C
+I_L3 = I_gC + I_f ; // Current delivered by dynamo C to the bus in A
+
+// case b
+// 1:
+P_LA = V_L * I_L1 ; // Power delivered to the bus by dynamo A in W
+P_gA = E_g1 * I_a1 ; // Power generated by dynamo A
+
+// 2:
+P_LB = V_L * I_L2 ; // Power delivered to the bus by dynamo B in W
+P_gB = E_g2 * I_a2 ; // Power generated by dynamo B
+
+// 3:
+P_LC = V_L * I_L3 ; // Power delivered to the bus by dynamo C in W
+P_gC = E_g3 * I_a3 ; // Power generated by dynamo C
+
+// Display the results
+disp("Example 7-1 Solution : ");
+printf(" \n a: 1. I_gA = %d A \t I_f = %d A ", I_gA,I_f );
+printf(" \n Thus,dynamo A delivers %d A to the bus and has an armature", I_gA);
+printf(" \n current of %d A + %d A = %d \n", I_gA,I_f,I_a1 );
+
+printf(" \n 2. I_gB = %d A ", I_gB);
+printf(" \n Thus, dynamo B is floating and has as armature & field current of %d A \n",I_f);
+
+printf(" \n 3. I_gC = %d A ",I_gC);
+printf(" \n Dynamo C receives %d A from the bus & has an armature current of %d A\n",I_L3,I_a3);
+
+printf(" \n b: 1. Power delivered to the bus by dynamo A is : ");
+printf(" \n P_LA = %d W ",P_LA);
+printf(" \n Power generated by dynamo A is \n P_gA = %d W \n",P_gA);
+
+printf(" \n 2. Since dynamo B neither delivers power to nor receives power from the bus, ");
+printf(" \n P_B = %d W ",P_LB);
+printf(" \n Power generated by dynamo B,to excite its field, is");
+printf(" \n P_gB = %d W \n ", P_gB);
+
+printf(" \n 3. Power delivered by the bus to dynamo C is ");
+printf(" \n P_LC = %d W ", P_LC);
+printf(" \n while the internal power delivered in the direction of rotation");
+printf(" \n of its prime mover to aid rotation is \n P_gC = %d W", P_gC );
diff --git a/1092/CH7/EX7.10/Example7_10.sce b/1092/CH7/EX7.10/Example7_10.sce new file mode 100755 index 000000000..2ee988e51 --- /dev/null +++ b/1092/CH7/EX7.10/Example7_10.sce @@ -0,0 +1,70 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_2_mag = 230 ; // Magnitude of voltage generated by alternator 2 in volt
+E_1_mag = 230 ; // Magnitude of voltage generated by alternator 1 in volt
+
+theta_2 = 180 ; // Phase angle of generated voltage by alternator 2 in degrees
+theta_1 = 20 ; // Phase angle of generated voltage by alternator 1 in degrees
+
+// writing given voltage in exponential form as follows
+// %pi/180 for degrees to radians conversion
+E_2 = E_2_mag * expm(%i * theta_2*(%pi/180) ); // voltage generated by alternator 2 in volt
+E_1 = E_1_mag * expm(%i * theta_1*(%pi/180) ); // voltage generated by alternator 1 in volt
+
+// writing given impedance(in ohm)in exponential form as follows
+Z_1 = 6 * expm(%i * 50*(%pi/180) ); // %pi/180 for degrees to radians conversion
+Z_2 = Z_1 ;
+Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees
+
+// Calculations
+E_r = E_2 + E_1 ; // Total voltage generated by Alternator 1 and 2 in volt
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case a
+I_s = E_r / (Z_1 + Z_2); // Synchronozing current in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// case b
+E_gp1 = E_1_mag;
+P_1 = E_gp1 * I_s_m * cosd(I_s_a - theta_1); // Synchronozing power developed by alternator 1 in W
+
+// case c
+E_gp2 = E_2_mag;
+P_2 = E_gp2 * I_s_m * cosd(I_s_a - theta_2); // Synchronozing power developed by alternator 2 in W
+
+// case d
+// but consider +ve vlaue for P_2 for finding losses, so
+P2 = abs(P_2);
+losses = P_1 - P2 ; // Losses in the armature in W
+
+// E_r_a yields -80 degrees which is equivalent to 100 degrees, so
+theta = 100 - I_s_a ; // Phase difference between E_r and I_s in degrees
+
+check = E_r_m * I_s_m * cosd(theta); // Verifying losses by Eq.7-7
+R_aT = 12*cosd(50) ; // total armature resistance of alternator 1 and 2 in ohm
+double_check = (I_s_m)^2 * (R_aT); // Verifying losses by Eq.7-7
+
+// Display the results
+disp("Example 7-10 Solution : ");
+printf(" \n a: I_s = ");disp(I_s);
+printf(" \n I_s = %.2f <%.2f A \n ",I_s_m, I_s_a );
+
+printf(" \n b: P_1 = %.f W (power delivered to bus)",P_1);
+printf(" \n Note:Slight variation in P_1 is due slight variations in ")
+printf(" \n phase angle of I_s,& angle btw (E_gp1,I_s)\n")
+printf(" \n P_2 = %.f W (power received from bus)\n",P_2);
+
+printf(" \n c: Losses: P_1 - P_2 = %.f W",losses);
+printf(" \n Check: E_a*I_s*cos(theta) = %.f W ",check );
+printf(" \n Double check : (I_s)^2*(R_a1+R_a2) = %.f W ",double_check );
diff --git a/1092/CH7/EX7.11/Example7_11.sce b/1092/CH7/EX7.11/Example7_11.sce new file mode 100755 index 000000000..e65ef2eea --- /dev/null +++ b/1092/CH7/EX7.11/Example7_11.sce @@ -0,0 +1,91 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// writing supply voltage in exponential form as follows
+// %pi/180 for degrees to radians conversion
+V_AB = 100 * expm(%i * 0*(%pi/180) ); // voltage supplied across A & B in volt
+V_BC = 100 * expm(%i * -120*(%pi/180) ); // voltage supplied across B & C in volt
+V_CA = 100 * expm(%i * 120*(%pi/180) ); // voltage supplied across C & A in volt
+
+disp("Example 7-11 : ");
+printf("\n Writing two mesh equations for I_1 and I_2 in fig.7-23a yields following\n array :");
+printf(" \n I_1 \t\t I_2 \t\t V ");
+printf(" \n ____________________________________________");
+printf(" \n 6 + j0 \t -3 + j0 \t 100 + j0 ");
+printf(" \n -3 + j0 \t 3 - j4 \t -50 - j86.6 ");
+
+// Calculations
+A = [ (6+%i*0) (-3+%i*0) ; (-3+%i*0) (3-%i*4) ]; // Matrix containing above mesh eqns array
+delta = det(A); // Determinant of A
+
+// case a
+I_1 = det( [ (100+%i*0) (-3+%i*0) ; (-50-%i*86.60) (3-%i*4) ] ) / delta ;
+// Mesh current I_1 in A
+I_1_m = abs(I_1);//I_1_m=magnitude of I_1 in A
+I_1_a = atan(imag(I_1) /real(I_1))*180/%pi;//I_1_a=phase angle of I_1 in degrees
+
+I_2 = det( [ (6+%i*0) (100+%i*0) ; (-3+%i*0) (-50-%i*86.6) ] ) / delta ;
+// Mesh current I_2 in A
+I_2_m = abs(I_2);//I_2_m=magnitude of I_2 in A
+I_2_a = atan(imag(I_2) /real(I_2))*180/%pi;//I_2_a=phase angle of I_2 in degrees
+
+// case b
+I_A = I_1 ; // Line current I_A in A
+I_A_m = abs(I_A);//I_A_m=magnitude of I_A in A
+I_A_a = atan(imag(I_A) /real(I_A))*180/%pi;//I_A_a=phase angle of I_A in degrees
+
+I_B = I_2 - I_1 ; // Line current I_B in A
+I_B_m = abs(I_B);//I_B_m=magnitude of I_B in A
+I_B_a = atan(imag(I_B) /real(I_B))*180/%pi - 180;//I_B_a=phase angle of I_B in degrees
+
+I_C = -I_2 ; // Line current I_C in A
+I_C_m = abs(I_C);//I_C_m=magnitude of I_C in A
+I_C_a = 180 + atan(imag(I_C) /real(I_C))*180/%pi;//I_C_a=phase angle of I_C in degrees
+
+// case c
+Z_A = 3 * expm(%i * 0*(%pi/180) ); // Impedance in line A in ohm
+Z_B = 3 * expm(%i * 0*(%pi/180) ); // Impedance in line B in ohm
+Z_C = 4 * expm(%i * -90*(%pi/180) ); // Impedance in line C in ohm
+
+V_AO = I_A * Z_A ; // Phase voltage V_AO in volt
+V_AO_m = abs(V_AO);//V_AO_m=magnitude of V_AO in volt
+V_AO_a = atan(imag(V_AO) /real(V_AO))*180/%pi;//V_AO_a=phase angle of V_AO in degrees
+
+V_BO = I_B * Z_B ; // Phase voltage V_BO in volt
+V_BO_m = abs(V_BO);//V_BO_m=magnitude of V_BO in volt
+V_BO_a = atan(imag(V_BO) /real(V_BO))*180/%pi - 180;//V_BO_a=phase angle of V_BO in degrees
+
+V_CO = I_C * Z_C ; // Phase voltage V_CO in volt
+V_CO_m = abs(V_CO);//V_CO_m=magnitude of V_CO in volt
+V_CO_a = atan(imag(V_CO) /real(V_CO))*180/%pi;//V_CO_a=phase angle of V_CO in degrees
+
+// Display the results
+disp("Solution : ");
+printf(" \n a: I_1 in A = ");disp(I_1);
+printf(" \n I_1 = %.2f <%.2f A \n ",I_1_m, I_1_a );
+printf(" \n I_2 in A = ");disp(I_2);
+printf(" \n I_2 = %.2f <%.2f A\n ",I_2_m, I_2_a );
+
+printf(" \n b: I_A in A = ");disp(I_1);
+printf(" \n I_A = %.2f <%.2f A\n",I_A_m, I_A_a );
+
+printf(" \n I_B in A = ");disp(I_B);
+printf(" \n I_B = %.2f <%.2f A\n",I_B_m, I_B_a );
+
+printf(" \n I_C in A = ");disp(I_C);
+printf(" \n I_C = %.2f <%.2f A \n",I_C_m, I_C_a );
+
+printf(" \n c: V_AO = %.2f <%.2f V",V_AO_m, V_AO_a );
+printf(" \n V_BO = %.2f <%.2f V",V_BO_m, V_BO_a );
+printf(" \n V_CO = %.2f <%.2f V\n",V_CO_m, V_CO_a );
+
+printf(" \n d: The phasor diagram is shown in Fig.7-23b, with the phase voltages");
+printf(" \n inscribed inside the (equilateral) triangle of given line voltages");
diff --git a/1092/CH7/EX7.2/Example7_2.sce b/1092/CH7/EX7.2/Example7_2.sce new file mode 100755 index 000000000..3f4111a97 --- /dev/null +++ b/1092/CH7/EX7.2/Example7_2.sce @@ -0,0 +1,40 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 7: PARALLEL OPERATION +// Example 7-2 + +clear; clc; close; // Clear the work space and console. + +// Given data +R_a = 0.1 ; // Armature resistance in ohm +R_f = 100 ; // Field ckt resistance in ohm +V_L_b = 120 ; // Bus voltage in volt +V_L_a = 140 ; // Voltage of the generator in volt +V_f = V_L_a ; // Voltage across the field in volt + +// Calculations +// case a +I_f_a = V_f / R_f ; // Field current in A +I_a_a = I_f_a ; // Armature current in A +E_g_a = V_L_a + I_a_a * R_a ; // Generated EMF in volt +P_g_a = E_g_a * I_a_a ; // Generated power in W + +// case b +I_a_b = ( E_g_a - V_L_b ) / R_a ; // Armature current in A +I_f_b = V_L_b / R_f ; // Field current in A +I_Lg = I_a_b - I_f_b ; // Generated line current in A +P_L = V_L_b * I_Lg ; // Power generated across the lines in W +E_g_b = V_L_a ; +P_g_b = E_g_b * I_a_b ; // Generated power in W + +// Display the results +disp("Example 7-2 Solution : "); +printf(" \n a: Before it is connected to the bus "); +printf(" \n I_a = I_f = %.1f A \n E_g = %.2f V\n P_g = %.1f W \n", I_a_a,E_g_a,P_g_a); + +printf(" \n b: After it is connected to the bus "); +printf(" \n I_a = %.1f A \n I_f = %.1f A \n I_Lg = %.1f A \n", I_a_b, I_f_b, I_Lg ); +printf(" \n P_L = %.f W \n P_g = %.f W ", P_L , P_g_b ); diff --git a/1092/CH7/EX7.3/Example7_3.sce b/1092/CH7/EX7.3/Example7_3.sce new file mode 100755 index 000000000..d08f6534a --- /dev/null +++ b/1092/CH7/EX7.3/Example7_3.sce @@ -0,0 +1,66 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_a = 0.1 ; // Armature resistance in ohm of 3 shunt generators
+R_a1 =R_a ;
+R_a2 =R_a ;
+R_a3 =R_a ;
+R_L = 2 ; // Load resistance in ohm
+E_g1 = 127 ; // Voltage generated by generator 1 in volt
+E_g2 = 120 ; // Voltage generated by generator 2 in volt
+E_g3 = 119 ; // Voltage generated by generator 3 in volt
+// Neglect field currents
+
+// Calculations
+// case a
+// Terminal bus voltage in volt
+V_L = ( (127/0.1) + (120/0.1) + (119/0.1) ) / ( (1/0.1) + (1/0.1) + (1/0.1) + 0.5);
+
+// case b
+I_L1 = (E_g1 - V_L)/R_a1 ; // Current delivered by generator 1 in A
+I_L2 = (E_g2 - V_L)/R_a2 ; // Current delivered by generator 2 in A
+I_L3 = (E_g3 - V_L)/R_a3 ; // Current delivered by generator 3 in A
+I_L_2ohm = V_L / R_L ; // Current delivered by 2 ohm load in A
+
+// case c
+I_a1 = I_L1 ; // Armature current in A for generator 1
+I_a2 = I_L2 ; // Armature current in A for generator 2
+I_a3 = I_L3 ; // Armature current in A for generator 3
+
+P_g1 = E_g1 * I_a1 ; // Power generated by generator 1 in W
+P_g2 = E_g2 * I_a2 ; // Power generated by generator 2 in W
+P_g3 = E_g3 * I_a3 ; // Power generated by generator 3 in W
+
+// case d
+P_L1 = V_L * I_L1 ; // Power delivered to or received from generator 1 in W
+P_L2 = V_L * I_L2 ; // Power delivered to or received from generator 2 in W
+P_L3 = V_L * I_L3 ; // Power delivered to or received from generator 3 in W
+P_L = V_L * -I_L_2ohm ; // Power delivered to or received 2 ohm load in W
+
+// Display the results
+disp("Example 7-3 Solution : ");
+printf(" \n a: Converting each voltage source to a current source and applying");
+printf(" \n Millman`s theorem yields ")
+printf(" \n V_L = %d V \n ", V_L );
+
+printf(" \n b: I_L1 = %d A (to bus)", I_L1 );
+printf(" \n I_L2 = %d A ", I_L2 );
+printf(" \n I_L3 = %d A (from bus)", I_L3 );
+printf(" \n I_L_2ohm = -%d A (from bus) \n", I_L_2ohm );
+
+printf(" \n c: P_g1 = %d W ",P_g1 );
+printf(" \n P_g2 = %d W (floating)",P_g2 );
+printf(" \n P_g3 = %d W \n",P_g3 );
+
+printf(" \n d: P_L1 = %d W ", P_L1);
+printf(" \n P_L2 = %d W ", P_L2 );
+printf(" \n P_L3 = %d W ", P_L3 );
+printf(" \n P_L = %d W ", P_L );
diff --git a/1092/CH7/EX7.4/Example7_4.sce b/1092/CH7/EX7.4/Example7_4.sce new file mode 100755 index 000000000..3269a85be --- /dev/null +++ b/1092/CH7/EX7.4/Example7_4.sce @@ -0,0 +1,46 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P1 = 300 ; // Power rating of generator 1 in kW
+P2 = 600 ; // Power rating of generator 2 in kW
+V = 220 ; // Voltage rating of generator 1 and 2 in volt
+V_o = 250 ; // No-load voltage applied to both the generators in volt
+// Assume linear characteristics
+V_1 = 230 ; // Terminal voltage in volt (case a)
+V_2 = 240 ; // Terminal voltage in volt (case b)
+
+// Calculations
+// case a
+kW1_a = (V_o - V_1)/(V_o - V) * P1 ; // kW carried by generator 1
+kW2_a = (V_o - V_1)/(V_o - V) * P2 ; // kW carried by generator 2
+
+// case b
+kW1_b = (V_o - V_2)/(V_o - V) * P1 ; // kW carried by generator 1
+kW2_b = (V_o - V_2)/(V_o - V) * P2 ; // kW carried by generator 2
+
+// case c
+frac_a = (V_o - V_1)/(V_o - V); // Fraction of rated kW carried by each generator
+frac_b = (V_o - V_2)/(V_o - V); // Fraction of rated kW carried by each generator
+
+// Display the results
+disp("Example 7-4 Solution : ");
+printf(" \n a: At 230 V, using Eq.(7-3) below : ");
+printf(" \n Generator 1 carries = %d kW ", kW1_a );
+printf(" \n Generator 2 carries = %d kW \n", kW2_a );
+
+printf(" \n b: At 240 V, using Eq.(7-3) below : ");
+printf(" \n Generator 1 carries = %d kW ", kW1_b );
+printf(" \n Generator 2 carries = %d kW \n", kW2_b );
+
+printf(" \n c: Both generators carry no-load at 250 V; ");
+printf(" \n %f rated load at %d V; ", frac_b , V_2 );
+printf(" \n %f rated load at %d V; ", frac_a , V_1 );
+printf(" \n and rated load at %d V. ", V );
diff --git a/1092/CH7/EX7.5/Example7_5.sce b/1092/CH7/EX7.5/Example7_5.sce new file mode 100755 index 000000000..393202959 --- /dev/null +++ b/1092/CH7/EX7.5/Example7_5.sce @@ -0,0 +1,38 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_1 = 220 ; // Terminal voltage of alternator 1 in volt
+E_2 = 222 ; // Terminal voltage of alternator 2 in volt
+f_1 = 60 ; // Frequency of alternator 1 in Hz
+f_2 = 59.5 ; // Frequency of alternator 2 in Hz
+// Switch is open
+
+// Calculations
+// case a
+E_max = (E_1 + E_2)/2 ; // Maximum effective voltage across each lamp in volt
+E_min = (E_2 - E_1)/2 ; // Minimum effective voltage across each lamp in volt
+
+// case b
+f = f_1 - f_2 ; // Frequency in Hz of the voltage across the lamps
+
+// case c
+E_peak = E_max / 0.7071 ; // Peak value of the voltage in volt across each lamp
+
+// case d
+n = (1/2)*f_1 ; // Number of maximum light pulsations per minute
+
+// Display the results
+disp("Example 7-5 Solution : ");
+printf(" \n a: E_max/lamp = %d V (rms)\n ", E_max );
+printf(" \n E_min/lamp = %d V \n ", E_min );
+printf(" \n b: f = %.1f Hz \n ", f );
+printf(" \n c: E_peak = %.f V \n ", E_peak );
+printf(" \n d: n = %d pulsations/min ", n );
diff --git a/1092/CH7/EX7.6/Example7_6.sce b/1092/CH7/EX7.6/Example7_6.sce new file mode 100755 index 000000000..6c1cf6a18 --- /dev/null +++ b/1092/CH7/EX7.6/Example7_6.sce @@ -0,0 +1,32 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E = 220 ; // Voltage generated in volt
+E_1 = E ; // Voltage generated by alternator 1 in volt
+E_2 = E ; // Voltage generated by alternator 2 in volt
+f_1 = 60 ; // Frequency in Hz of alternator 1
+f_2 = 58 ; // Frequency in Hz of alternator 2
+// Switch is open
+
+// Calculations
+// case a
+E_max = (E_1 + E_2)/2 ; // Maximum effective voltage across each lamp in volt
+f = f_1 - f_2 ; // Frequency in Hz of the voltage across the lamps
+
+// case c
+E_min = (E_2 - E_1)/2 ; // Minimum effective voltage across each lamp in volt
+
+// Display the results
+disp("Example 7-6 Solution : ");
+printf(" \n a: E_max/lamp = %d V \n f = %d Hz \n ", E_max, f );
+printf(" \n b: The voltages are equal and opposite in the local circuit. \n ");
+printf(" \n c: E_min/lamp = %d V at zero frequency \n ", E_min );
+printf(" \n d: The voltages are in phase in the local circuit. ");
diff --git a/1092/CH7/EX7.7/Example7_7.sce b/1092/CH7/EX7.7/Example7_7.sce new file mode 100755 index 000000000..7e08faf5b --- /dev/null +++ b/1092/CH7/EX7.7/Example7_7.sce @@ -0,0 +1,57 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data as per Ex.(7-5)
+E1 = 220 ; // Terminal voltage of alternator 1 in volt
+E2 = 222 ; // Terminal voltage of alternator 2 in volt
+f1 = 60 ; // Frequency of alternator 1 in Hz
+f2 = 59.5 ; // Frequency of alternator 2 in Hz
+// Switch is open
+
+// Given data as per Ex.(7-6)
+E = 220 ; // Voltage generated in volt
+E_1 = E ; // Voltage generated by alternator 1 in volt
+E_2 = E ; // Voltage generated by alternator 2 in volt
+f_1 = 60 ; // Frequency in Hz of alternator 1
+f_2 = 58 ; // Frequency in Hz of alternator 2
+// Switch is open
+
+// Given data as per Ex.(7-7)
+R_a1 = 0.1 ; // armature resistance of alternator 1 in ohm
+R_a2 = 0.1 ; // armature resistance of alternator 2 in ohm
+X_a1 = 0.9 ; // armature reactance of alternator 1 in ohm
+X_a2 = 0.9 ; // armature reactance of alternator 2 in ohm
+
+Z_1 = R_a1 + %i*X_a1 ; // Effective impedance of alternator 1 in ohm
+Z_2 = R_a1 + %i*X_a2 ; // Effective impedance of alternator 2 in ohm
+// Switches are closed at the proper instant for paralleling.
+
+// Calculations
+// In Ex.7-5,
+E_r = E2 - E1 ; // Effective voltage generated in volt
+I_s = E_r / (Z_1 + Z_2); // Synchronizing current in the armature in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// In Ex.7-6,
+Er = E_2 -E_1 ; // Effective voltage generated in volt
+Is = Er / ( Z_1 + Z_2); // Synchronizing current in the armature in A
+
+// Display the results
+disp("Example 7-7 Solution : ");
+printf(" \n In Ex.7-5, ");
+printf(" \n E_r = %d V ", E_r);
+printf(" \n I_s = ");disp(I_s);
+printf(" \n I_s = %.3f <%.2f A ",I_s_m, I_s_a);
+printf(" \n where %.3f is magnitude in A and %.2f is phase angle in degrees \n",I_s_m,I_s_a);
+
+printf(" \n In Ex.7-6, ");
+printf(" \n E_r = %d V ", Er );
+printf(" \n I_s = %d A",Is);
diff --git a/1092/CH7/EX7.8/Example7_8.sce b/1092/CH7/EX7.8/Example7_8.sce new file mode 100755 index 000000000..fc8a8afc0 --- /dev/null +++ b/1092/CH7/EX7.8/Example7_8.sce @@ -0,0 +1,71 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// EMF's are opposed exactly by 180 degrees
+E_gp1 = 200 ; // Terminal voltage of alternator 1 in volt
+E_gp2 = 220 ; // Terminal voltage of alternator 2 in volt
+R_a1 = 0.2 ; // armature resistance of alternator 1 in ohm
+R_a2 = 0.2 ; // armature resistance of alternator 2 in ohm
+X_a1 = 2 ; // armature reactance of alternator 1 in ohm
+X_a2 = 2 ; // armature reactance of alternator 2 in ohm
+
+Z_p1 = R_a1 + %i*X_a1 ; // Effective impedance of alternator 1 in ohm
+Z_p2 = R_a1 + %i*X_a2 ; // Effective impedance of alternator 2 in ohm
+// Switches are closed at the proper instant for paralleling.
+
+// Calculations
+// case a
+E_r = (E_gp2 - E_gp1) ; // Effective voltage generated in volt
+I_s = E_r / (Z_p1 + Z_p2); // Synchronizing current in the armature in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+P_2 = E_gp2 * I_s_m * cosd(I_s_a); // Generator action developed by alternator 2 in W
+
+// case b
+theta = I_s_a;
+// P_1 = E_gp1 * I_s_m * cosd(180 - theta)
+// P_1 = -E_gp1 * I_s_m * cosd(theta),
+P_1 = -E_gp1 * I_s_m * cosd(theta); // Synchronizing power received by alternator 1 in W
+
+// case c
+// but consider +ve vlaue for P_1 for finding losses, so
+P1 = abs(P_1);
+losses = P_2 - P1 ; // Power losses in both armatures in W
+check = E_r * I_s_m * cosd(I_s_a); // Verifying losses by Eq.7-7
+double_check = (I_s_m)^2 * (R_a1 + R_a2); // Verifying losses by Eq.7-7
+
+// case d
+V_p2 = E_gp2 - I_s*Z_p1 ; // Generator action
+V_p1 = E_gp1 + I_s*Z_p1 ; // Motor action
+
+// Display the results
+disp("Example 7-8 Solution : ");
+printf(" \n a: E_r = %d V ",E_r);
+printf(" \n I_s = %.2f <%.2f A ", I_s_m, I_s_a );
+printf(" \n P_2 = %.1f W (total power delivered by alternator 2 ) \n", P_2);
+
+printf(" \n b: P_1 = %f W (synchronizing power received by alternator 1)",P_1);
+printf(" \n Note:Scilab considers phase angle of I_s as %f instead ",I_s_a);
+printf(" \n of -84.3 degrees,so slight variation in the answer P_1.\n");
+
+printf(" \n c: Consider +ve value of P_1 for calculating losses");
+printf(" \n Losses: P_2 - P_1 = %.1f W ",losses );
+printf(" \n Check: E_a*I_s*cos(theta) = %.1f W ",check );
+printf(" \n Double check : (I_s)^2*(R_a1+R_a2) = %.1f W as given in Eq.(7-1)",double_check );
+
+printf("\n\n d: From Fig.7-14, V_p2, the terminal phase voltage of ");
+printf(" \n alternator 2, is, from Eq.(7-1)");
+printf(" \n V_p2 = %d V (generator action)\n\n From section 7-2.1 ",V_p2);
+printf(" \n V_p1 = %d V ( motor action)\n",V_p1);
+
+printf(" \n e: The phasor diagram is shown in Fig.7-14.");
+
diff --git a/1092/CH7/EX7.9/Example7_9.sce b/1092/CH7/EX7.9/Example7_9.sce new file mode 100755 index 000000000..d2aaf5290 --- /dev/null +++ b/1092/CH7/EX7.9/Example7_9.sce @@ -0,0 +1,73 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_2_mag = 230 ; // Magnitude of voltage generated by alternator 2 in volt
+E_1_mag = 230 ; // Magnitude of voltage generated by alternator 1 in volt
+
+theta_2 = 180 ; // Phase angle of generated voltage by alternator 2 in degrees
+theta_1 = 20 ; // Phase angle of generated voltage by alternator 1 in degrees
+
+R_a1 = 0.2 ; // armature resistance of alternator 1 in ohm
+R_a2 = 0.2 ; // armature resistance of alternator 2 in ohm
+
+// writing given voltage in exponential form as follows
+// %pi/180 for degrees to radians conversion
+E_2 = E_2_mag * expm(%i * theta_2*(%pi/180) ); // voltage generated by alternator 2 in volt
+E_1 = E_1_mag * expm(%i * theta_1*(%pi/180) ); // voltage generated by alternator 1 in volt
+
+// writing given impedance(in ohm)in exponential form as follows
+Z_1 = 2.01 * expm(%i * 84.3*(%pi/180) ); // %pi/180 for degrees to radians conversion
+Z_2 = Z_1 ;
+Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees
+
+// Calculations
+E_r = E_2 + E_1 ; // Total voltage generated by Alternator 1 and 2 in volt
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case a
+I_s = E_r / (Z_1 + Z_2); // Synchronozing current in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// case b
+E_gp1 = E_1_mag;
+P_1 = E_gp1 * I_s_m * cosd(I_s_a - theta_1); // Synchronozing power developed by alternator 1 in W
+
+// case c
+E_gp2 = E_2_mag;
+P_2 = E_gp2 * I_s_m * cosd(I_s_a - theta_2); // Synchronozing power developed by alternator 2 in W
+
+// case d
+// but consider +ve vlaue for P_2 for finding losses, so
+P2 = abs(P_2);
+losses = P_1 - P2 ; // Losses in the armature in W
+
+// E_r_a yields -80 degrees which is equivalent to 100 degrees, so
+theta = 100 - I_s_a ; // Phase difference between E_r and I_a in degrees
+
+check = E_r_m * I_s_m * cosd(theta); // Verifying losses by Eq.7-7
+R_aT = R_a1 + R_a2 ; // total armature resistance of alternator 1 and 2 in ohm
+double_check = (I_s_m)^2 * (R_aT); // Verifying losses by Eq.7-7
+
+// Display the results
+disp("Example 7-9 Solution : ");
+printf(" \n a: I_s = ");disp(I_s);
+printf(" \n I_s = %.2f <%.2f A \n ",I_s_m, I_s_a );
+
+printf(" \n b: P_1 = %.f W (power delivered to bus)",P_1);
+printf(" \n Slight variation in P_1 is due slight variations in ")
+printf(" \n magnitude of I_s,& angle btw (E_gp1,I_s)\n")
+printf(" \n P_2 = %.f W (power received from bus)\n",P_2);
+
+printf(" \n c: Losses: P_1 - P_2 = %d",losses);
+printf(" \n Check: E_a*I_s*cos(theta) = %d W ",check );
+printf(" \n Double check : (I_s)^2*(R_a1+R_a2) = %d W ",double_check );
diff --git a/1092/CH8/EX8.1/Example8_1.sce b/1092/CH8/EX8.1/Example8_1.sce new file mode 100755 index 000000000..25d7c9dcf --- /dev/null +++ b/1092/CH8/EX8.1/Example8_1.sce @@ -0,0 +1,72 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3- phase Y-connected synchronous motor
+P = 20 ; // No. of poles
+hp = 40 ; // power rating of the synchronous motor in hp
+V_L = 660 ; // Line voltage in volt
+beta = 0.5 ; // At no-load, the rotor is retarded 0.5 mechanical degree from
+// its synchronous position.
+X_s = 10 ; // Synchronous reactance in ohm
+R_a = 1.0 ; // Effective armature resistance in ohm
+
+// Calculations
+// case a
+funcprot(0); // To avoid this message "Warning : redefining function: beta"
+alpha = P * (beta/2); // The rotor shift from the synchronous position in
+// electrical degrees.
+
+// case b
+V_p = V_L / sqrt(3); // Phase voltage in volt
+E_gp = V_p ; // Generated voltage/phase at no-load in volt (given)
+E_r = (V_p - E_gp*cosd(alpha)) + %i*(E_gp*sind(alpha));
+ // Resultant emf across the armature per phase in V/phase
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case c
+Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm
+Z_s_m = abs(Z_s);//Z_s_m=magnitude of Z_s in ohm
+Z_s_a = atan(imag(Z_s) /real(Z_s))*180/%pi;//Z_s_a=phase angle of Z_s in degrees
+
+I_a = E_r / Z_s ; // Armature current/phase in A/phase
+I_a_m = abs(I_a);//I_a_m=magnitude of I_a in A
+I_a_a = atan(imag(I_a) /real(I_a))*180/%pi;//I_a_a=phase angle of I_a in degrees
+
+// case d
+theta = I_a_a ; // Phase angle between V_p and I_a in degrees
+P_p = V_p * I_a_m * cosd(theta); // Power per phase drawn by the motor from the bus
+P_t = 3*P_p ; // Total power drawn by the motor from the bus
+
+// csae e
+P_a = 3 * (I_a_m)^2 * R_a ; // Armature power loss at no-load in W
+P_d = (P_t - P_a)/746 ; // Internal developed horsepower at no-load
+
+// Display the results
+disp("Example 8-1 Solution : ");
+printf(" \n a: alpha = %d degrees (electrical degrees)\n",alpha );
+
+printf(" \n b: E_gp = %d V also, as given ",E_gp);
+printf(" \n E_r in V/phase = ");disp(E_r);
+printf(" \n E_r = %.1f <%.1f V/phase \n",E_r_m, E_r_a );
+
+printf(" \n c: Z_s in ohm/phase = ");disp(Z_s);
+printf(" \n Z_s = %.2f <%.1f ohm/phase \n",Z_s_m, Z_s_a );
+printf(" \n I_a in A/phase = ");disp(I_a);
+printf(" \n I_a = %.2f <%.2f A/phase \n ",I_a_m, I_a_a);
+
+printf(" \n d: P_p = %.2f W/phase ",P_p );
+printf(" \n P_t = %.2f W ",P_t);
+printf(" \n Note: Slight variations in power values is due to slight variations");
+printf(" \n in V_p , I_a and theta values from those of the textbook\n");
+
+printf(" \n e: P_a = %.f W ",P_a );
+printf(" \n P_d = %d hp ", P_d );
diff --git a/1092/CH8/EX8.10/Example8_10.sce b/1092/CH8/EX8.10/Example8_10.sce new file mode 100755 index 000000000..dfdd17198 --- /dev/null +++ b/1092/CH8/EX8.10/Example8_10.sce @@ -0,0 +1,56 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS +// Example 8-10 + +clear; clc; close; // Clear the work space and console. + +// Given data +kVA = 10000 ; // kVA rating of a system +cos_theta = 0.65 ; // power factor of the system +sin_theta = sqrt( 1 - (cos_theta)^2 ); +cos_theta_b = 0.85 ; // Raised PF +sin_theta_b = sqrt( 1 - (cos_theta_b)^2 ); +cost = 60 ; // cost of the synchronous capacitor to improve the PF in dollars/kVA +// neglect the losses in the synchronous capacitor + +// Calculations +// case a : For unity PF +// at the original load +kW_a = kVA * cos_theta ; // +theta = acosd(cos_theta) ; // Power factor angle of the system in degrees +kvar = kVA * sind(theta) ; // Reactive power in kvar +kVA_a = kvar ; +cost_cap_a = kvar * cost ; // Cost of raising the PF to unity PF in dollars + +// case b +theta_b = acosd(cos_theta_b) ; // Power factor angle of the system in degrees +kVA_b = kW_a / cos_theta_b ; // kVA value reduction +kvar_b = kVA_b * sind(theta_b) ; // final kvar value reduced +kvar_add = kvar - kvar_b ; // kvar of correction added + +cost_cap_b = kvar_add * cost ; // Cost of raising the PF to 0.85 PF in dollars + +// Display the results + +disp("Example 8-10 Solution : "); +printf(" \n Note : Slight variations in the kvar and cost values are due to "); +printf(" \n non-approximation of theta values while calculating in scilab.\n"); +printf(" \n a: At the original load,\n"); +printf(" \n kW = %d kW at theta = %.1f degrees \n", kW_a , theta ); +printf(" \n kvar = %.3f kvar\n\n For unity PF,",kvar); +printf(" \n kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n",kVA_a); +printf(" \n Cost of synchronous capacitor = $%.f \n\n",cost_cap_a ); + +printf(" \n b: For %.2f, PF = cos(%.1f), the total power,",cos_theta_b, theta_b); +printf(" \n %.f kW,remains the same. Therefore,\n ",kW_a); +printf(" \n kVA of final system reduced to = %.f kVA \n",kVA_b); +printf(" \n kvar of final system reduced to = %.f kvar \n Therefore,",kvar_b); + +printf(" \n kvar of correction added = %.3f kvar\n ",kvar_add); +printf(" \n kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n",kvar_add); +printf(" \n Cost of synchronous capacitor = $%.f",cost_cap_b ); +printf(" \n or less than half the cost in part(a)"); diff --git a/1092/CH8/EX8.11/Example8_11.sce b/1092/CH8/EX8.11/Example8_11.sce new file mode 100755 index 000000000..ed54ffbd6 --- /dev/null +++ b/1092/CH8/EX8.11/Example8_11.sce @@ -0,0 +1,31 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S_conjugate = 1000 ; // Apparent complex power in kVA
+cos_theta = 0.6 ; // lagging PF
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+
+// Calculations
+// case a
+P_o = S_conjugate * cos_theta ; // Active power dissipated by the load in kW
+
+// case b
+jQ_o = S_conjugate * sin_theta ; // Inductive reactive quadrature power -
+// - drawn from and returned to the supply
+
+// Display the results
+
+disp("Example 8-11 Solution : ");
+printf(" \n a: Active power \n P_o = %d kW \n ", P_o );
+
+printf(" \n b: Inductive reactive quadrature power \n +jQ_o in kvar = \n");disp(%i*jQ_o);
+
+printf(" \n c: The original power triangle is shown in Fig.8-26a.");
diff --git a/1092/CH8/EX8.12/Example8_12.sce b/1092/CH8/EX8.12/Example8_12.sce new file mode 100755 index 000000000..6efcd28af --- /dev/null +++ b/1092/CH8/EX8.12/Example8_12.sce @@ -0,0 +1,49 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-12
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+S_conjugate = 1000 ; // Apparent complex power in kVA
+cos_theta_f = 0.8 ; // lagging PF
+sin_theta_f = sqrt( 1 - (cos_theta_f)^2 );
+
+// Calculated values from Ex.8-11
+P_o = 600 ; // Active power dissipated by the load in kW
+Q_o = 800 ; // Inductive reactive quadrature power -
+// - drawn from and returned to the supply
+
+// Calculations :
+
+// case a
+P_f = S_conjugate * cos_theta_f ; // Active power dissipated by the load in kW
+
+// case b
+Q_f = S_conjugate * sin_theta_f ; // Reactive quadrature power drawn from -
+// - and returned to the supply
+
+// case c
+P_a = P_f - P_o ; // Additional active power in kW that may be supplied to -
+// - new customers
+
+// case d
+jQ_a = %i * ( Q_f ) - %i * ( Q_o ); // Correction kvar required to raise PF -
+// -from 0.6 to o.8 lagging
+
+// case e
+S_c_conjugate = 0 - jQ_a ; // Rating of correction capacitors needed for case d
+
+// Display the results
+
+disp("Example 8-12 Solution : ");
+printf(" \n a: P_f = %d kW \n ", P_f );
+printf(" \n b: +jQ_f in kvar = ");disp(%i*Q_f);
+printf(" \n c: P_a = %d kW \n ", P_a );
+printf(" \n d: jQ_a in kvar = ");disp(jQ_a)
+printf(" \n e: S_c_conjugate = %d kVA \n ", abs(S_c_conjugate) );
+printf(" \n f: The power tabulation grid is shown in Fig.8-26b.");
diff --git a/1092/CH8/EX8.13/Example8_13.sce b/1092/CH8/EX8.13/Example8_13.sce new file mode 100755 index 000000000..3ddf06d9b --- /dev/null +++ b/1092/CH8/EX8.13/Example8_13.sce @@ -0,0 +1,59 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-13
+
+clear; clc; close; // Clear the work space and console.
+
+// Ex.8-12 PF
+cos_theta = 0.6 ; // PF lagging
+
+// Given data
+S_conjugate = 1000 ; // Apparent complex power in kVA
+cos_theta_f = 1.0 ; // unity PF
+sin_theta_f = sqrt( 1 - (cos_theta_f)^2 );
+
+// Calculated values from Ex.8-11
+P_o = 600 ; // Active power dissipated by the load in kW
+Q_o = 800 ; // Inductive reactive quadrature power -
+// - drawn from and returned to the supply
+
+// Calculations :
+
+// case a
+P_f = S_conjugate * cos_theta_f ; // Active power dissipated by the load in kW
+
+// case b
+Q_f = S_conjugate * sin_theta_f ; // Reactive quadrature power drawn from -
+// - and returned to the supply
+
+// case c
+P_a = P_f - P_o ; // Additional active power in kW that may be supplied to -
+// - new customers
+
+// case d
+jQ_a = %i * ( Q_f ) - %i * ( Q_o ); // Correction kvar required to raise PF -
+// -from 0.6 to o.8 lagging
+Q_a = -abs(jQ_a); //
+
+// case e
+S_c_conjugate = 0 - jQ_a ; // Rating of correction capacitors needed for case d
+
+// Display the results
+
+disp("Example 8-13 Solution : ");
+printf(" \n a: P_f = %d kW \n ", P_f );
+printf(" \n b: +jQ_f in kvar = ");disp(%i*Q_f);
+printf(" \n c: P_a = %d kW \n ", P_a );
+printf(" \n d: jQ_a in kvar = ");disp(jQ_a)
+printf(" \n e: S_c_conjugate = %d kVA \n ", abs(S_c_conjugate) );
+printf(" \n f: The power tabulation grid is shown below.\n");
+printf(" \n \t\t P \t ±jQ \t S* ");
+printf(" \n \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ");
+printf(" \n ___________________________________________");
+printf(" \n Original : \t %d \t +j%d \t %d \t %.1f ",P_o ,Q_o ,S_conjugate,cos_theta);
+printf(" \n Added : \t %d \t %dj \t __ \t __",P_a ,Q_a );
+printf(" \n Final : \t %d \t +j%d \t %d \t %.1f",P_f ,Q_f ,S_conjugate,cos_theta_f);
diff --git a/1092/CH8/EX8.14/Example8_14.sce b/1092/CH8/EX8.14/Example8_14.sce new file mode 100755 index 000000000..abb791b02 --- /dev/null +++ b/1092/CH8/EX8.14/Example8_14.sce @@ -0,0 +1,69 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-14
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2000 ; // load in kW drawn by a factory
+cos_theta_o = 0.6 ; // PF lagging
+sin_theta_o = sqrt( 1- (cos_theta_o)^2 );
+cos_theta_f = 0.85 ; // final PF lagging required
+sin_theta_f = sqrt( 1- (cos_theta_f)^2 );
+P_a = 275 ; // Losses in the synchronous capacitor in kW
+
+// Calculations
+// case a
+S_o_conjugate = P_o / cos_theta_o ; // Original kVA drawn from the utility
+
+// case b
+Q_o = S_o_conjugate * sin_theta_o ; // Original lagging kvar
+
+// case c
+P_f = P_o + P_a ; // Final system active power consumed from the utility in kW
+
+// case d
+S_f_conjugate = P_f / cos_theta_f ; // Final kVA drawn from the utility
+S_f_conjugate_a = acosd(cos_theta_f); // Phase angle of S_f_conjugate in degrees
+
+// case e
+jQ_f = S_f_conjugate * sin_theta_f ; // Final lagging kvar
+jQ_a = %i*(jQ_f) - %i*(Q_o); // Correction kvar produced by the synchronous capacitor
+Q_a = abs(jQ_a); // Magnitude of jQ_a in kvar
+
+// case f
+P = P_a ;
+S_a_conjugate = P -%i*(abs(jQ_a)); // kVA rating of the synchronous capacitor
+S_a_conjugate_m = abs(S_a_conjugate);//S_a_conjugate_m = magnitude of S_a_conjugate in kVA
+S_a_conjugate_a = atan(imag(S_a_conjugate) /real(S_a_conjugate))*180/%pi;
+//S_a_conjugate_a=phase angle of S_a_conjugate in degrees
+PF_f = cosd(S_a_conjugate_a); // PF
+
+// Display the results
+disp("Example 8-14 Solution : ");
+printf(" \n a: S*o = %.1f kVA \n",S_o_conjugate );
+
+printf(" \n b: Q*o in kvar = " );disp(%i*Q_o);
+
+printf(" \n c: P*f = %.f kW \n",P_f );
+
+printf(" \n d: S*f = %.1f <%.1f kVA\n ",S_f_conjugate,S_f_conjugate_a );
+
+printf(" \n e: jQ_f in kvar = ");disp(%i*jQ_f);
+printf(" \n -jQ_a in kvar = ");disp(jQ_a);
+
+printf(" \n f: S*a = %.f <%.2f kVA ", S_a_conjugate_m , S_a_conjugate_a );
+printf(" \n (cos(%.2f) = %.3f leading)\n",S_a_conjugate_a,PF_f);
+
+printf(" \n g: Power tabulation grid : \n ");
+printf(" \n \t\t P \t ±jQ \t S* ");
+printf(" \n \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ");
+printf(" \n ___________________________________________");
+printf(" \n Original : \t %d \t +j%.f %.1f %.1f lag",P_o ,Q_o ,S_o_conjugate,cos_theta_o);
+printf(" \n Added : \t %d \t -%.fj %.f \t %.3f lead",P_a ,Q_a,S_a_conjugate_m,cosd(S_a_conjugate_a) );
+printf(" \n Final : \t %d \t +j%.f %.1f %.2f lag",P_f ,jQ_f ,S_f_conjugate,cos_theta_f);
+
diff --git a/1092/CH8/EX8.15/Example8_15.sce b/1092/CH8/EX8.15/Example8_15.sce new file mode 100755 index 000000000..d598f602a --- /dev/null +++ b/1092/CH8/EX8.15/Example8_15.sce @@ -0,0 +1,87 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-15
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2275 ; // Original kVA
+Q_o = 1410 ; // Original kvar
+S_f_conjugate = 3333.3 ; // final kVA of the load
+S_o_conjugate = P_o + %i*Q_o ; // Load of the alternator in kVA
+S_o_conjugate_m = abs(S_o_conjugate);//S_o_conjugate_m = magnitude of S_o_conjugate in kVA
+S_o_conjugate_a = atan(imag(S_o_conjugate) /real(S_o_conjugate))*180/%pi;
+//S_o_conjugate_a=phase angle of S_o_conjugate in degrees
+
+disp("Example 8-15");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n ________________________________________________________________________");
+printf(" \n Original : \t%d \t\t j%.f \t\t %.1f \t%.2f lag",real(S_o_conjugate) ,imag(S_o_conjugate) ,S_o_conjugate_m,cosd(S_o_conjugate_a));
+printf(" \n Added : \t0.8x \t\t j0.6x \t\t x \t\t0.80 lag" );
+printf(" \n Final : (%d + 0.8x) \tj(%.f + 0.6x) %.1f \t0.841 lag\n",real(S_o_conjugate) ,imag(S_o_conjugate),S_f_conjugate );
+
+// Calculations
+// case a
+// Assume x is the additional kVA load. Then real and quadrature powers are 0.8x and j0.6x
+// respectively,as shown. Adding each column vertically and using the Pythagorean theorem,
+// we may write (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this eqution yields
+// the quadratic x^2 + 5352x -3947163 = 0. Applying the quadratic yields the added kVA load:
+x = poly(0,'x'); // Defining a polynomial with variable 'x' with root at 0
+p = -3947163 + 5352*x + x^2
+a = 1 ; // coefficient of x^2
+b = 5332 ; // coefficient of x
+c = -3947163 ; // constant
+
+// Roots of p
+x1 = ( -b + sqrt (b^2 -4*a*c ) ) /(2* a);
+x2=( -b - sqrt (b^2 -4*a*c ) ) /(2* a);
+
+// case b
+P_a = 0.8*x1 ; // Added active power of the additional load in kW
+Q_a = 0.6*x1 ; // Added reactive power of the additional load in kvar
+
+// case c
+P_f = P_o + P_a ; // Final active power of the additional load in kW
+Q_f = Q_o + Q_a ; // Final reactive power of the additional load in kvar
+
+// case d
+PF = P_f / S_f_conjugate ; // Final power factor
+// Validity check
+S_conjugate_f = P_f + %i*Q_f ; // Final kVA of the load
+S_conjugate_f_m = abs(S_conjugate_f);//S_conjugate_f_m = magnitude of S_conjugate_f in kVA
+S_conjugate_f_a = atan(imag(S_conjugate_f) /real(S_conjugate_f))*180/%pi;
+//S_conjugate_f_a=phase angle of S_conjugate_f in degrees
+
+// Display the results
+
+disp(" Solution : ")
+
+printf(" \n a: The given data is shown in the above power tabulation grid.Assume");
+printf(" \n x is the additional kVA load. Then real and quadrature powers are");
+printf(" \n 0.8x and j0.6x respectively,as shown.Adding each column vertically");
+printf(" \n and using the Pythagorean theorem, we may write");
+printf(" \n (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this");
+printf(" \n equation yields the quadratic as follows : \n");
+printf(" \n x^2 + 5332x -3947163 = 0. \n ")
+printf(" \n Applying the quadratic yields the added kVA load:");
+printf(" \n Roots of quadratic Eqn p are \n ");
+printf(" \n x1 = %.2f \n x2 = %.2f ", x1, x2 );
+printf(" \n Consider +ve value of x for added kVA so");
+printf(" \n x = S*a = %.2f kVA \n ", x1 );
+
+printf(" \n b: P_a = %.1f kW \n ", P_a );
+printf(" \n Q_a in kvar = \n");disp(%i*Q_a);
+
+printf(" \n c: P_f = %.1f kW \n ", P_f );
+printf(" \n Q_f in kvar = \n");disp(%i*Q_f);
+
+printf(" \n d: PF = cosθ_f = %.3f lagging \n ", PF );
+printf(" \n Validity check\n S*f = ");disp(S_conjugate_f);
+printf(" \n S*f = %.1f <%.2f kVA \n",S_conjugate_f_m,S_conjugate_f_a);
+printf(" \n PF = cos(%.1f) = %.3f lagging",S_conjugate_f_a ,cosd(S_conjugate_f_a));
diff --git a/1092/CH8/EX8.16/Example8_16.sce b/1092/CH8/EX8.16/Example8_16.sce new file mode 100755 index 000000000..37ef37246 --- /dev/null +++ b/1092/CH8/EX8.16/Example8_16.sce @@ -0,0 +1,45 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-16
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Calculated values as per Ex.8-15 are as follows
+S_o_conjugate = 2676.5*exp(%i*31.79*(%pi/180)); // Original kVA rating
+S_o_conjugate_m = abs(S_o_conjugate);//S_o_conjugate_m = magnitude of S_o_conjugate in kVA
+S_o_conjugate_a = atan(imag(S_o_conjugate) /real(S_o_conjugate))*180/%pi;
+//S_o_conjugate_a=phase angle of S_o_conjugate in degrees
+
+S_a_conjugate = 658.86*exp(%i*36.87*(%pi/180)); // Added kVA rating
+S_a_conjugate_m = abs(S_a_conjugate);//S_a_conjugate_m = magnitude of S_a_conjugate in kVA
+S_a_conjugate_a = atan(imag(S_a_conjugate) /real(S_a_conjugate))*180/%pi;
+//S_a_conjugate_a=phase angle of S_a_conjugate in degrees
+
+S_f_conjugate = -3333.3*exp(%i*32.792687*(%pi/180)); // Final kVA rating
+S_f_conjugate_m = abs(S_f_conjugate);//S_f_conjugate_m = magnitude of S_f_conjugate in kVA
+S_f_conjugate_a = atan(imag(S_f_conjugate) /real(S_f_conjugate))*180/%pi;
+//S_f_conjugate_a=phase angle of S_f_conjugate in degrees
+
+// Calculations
+kVA_total = S_o_conjugate + S_a_conjugate + S_f_conjugate ; // Tellegan's theorem
+kVA_total_m = abs(kVA_total);//kVA_total_m = magnitude of kVA_total in kVA
+kVA_total_a = atan(imag(kVA_total) /real(kVA_total))*180/%pi;
+//kVA_total_a=phase angle of kVA_total in degrees
+
+// Display the result
+disp("Example 8-16 Solution : ");
+printf(" \n From the solution to Ex.8-15, we have ");
+printf(" \n S*o = %.1f <%.2f kVA \n ", S_o_conjugate_m,S_o_conjugate_a );
+printf(" \n S*a = %.1f <%.2f kVA \n ", S_a_conjugate_m,S_a_conjugate_a );
+printf(" \n S*f = %.1f <%.2f kVA \n ", S_f_conjugate_m,S_f_conjugate_a );
+
+printf(" \n Validity check ");
+printf(" \n S*o + S*a + S*f = ");
+disp(S_o_conjugate),printf(" +"),disp(S_a_conjugate),printf(" +"),disp(S_f_conjugate);
+printf(" \n = %d ",kVA_total );
+printf(" \n Hence, Tellegen`s theorem is proved");
diff --git a/1092/CH8/EX8.17/Example8_17.sce b/1092/CH8/EX8.17/Example8_17.sce new file mode 100755 index 000000000..04e50959c --- /dev/null +++ b/1092/CH8/EX8.17/Example8_17.sce @@ -0,0 +1,68 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-17
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 40000 ; // Load on a factory in kW
+PF = 0.8 ; // power factor lagging of the load
+cos_theta = PF;
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+hp = 7500 ; // power rating of the induction motor in hp
+PF_IM = 0.75 ; // power factor lagging of the induction motor
+cos_theta_IM = PF_IM;
+sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );
+eta = 91*(1/100) ; // Efficiency of IM
+PF_SM = 1 ; // power factor of the synchronous motor
+
+// Calculations
+kVA_original = kW / PF ; // Original kVA
+kvar_original = kVA_original * sin_theta ; // Original kvar
+
+kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW
+kVA_IM = kW_IM / PF_IM ; // Induction motor kVA
+kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar
+
+kvar_final = kvar_original - kvar_IM ; // final kvar
+kVA_final = kW + %i*(abs(kvar_final)); // final kVA
+kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA
+kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;
+//kVA_final_a=phase angle of kVA_final in degrees
+
+PF_final = cosd(kVA_final_a); // Final power factor
+
+// Display the result
+disp("Example 8-17 Solution : ");
+printf(" \n The synchronous motor operates at the same efficiency as the IM");
+printf(" \n that has been replaced, and therefore the total power of the system");
+printf(" \n is unchanged. The solution involves construction of table that shows ")
+printf(" \n the original condition of the system, the change, and the final condition.\n");
+printf(" \n Original kVA = %d kVA \n ", kVA_original );
+printf(" \n Original kvar = \n" );disp(%i*kvar_original);
+
+printf(" \n Induction motor kW = %d kW \n ", kW_IM );
+printf(" \n Induction motor kVA = %.f kVA \n ", kVA_IM );
+printf(" \n Induction motor kvar = ");disp(%i*kvar_IM)
+
+printf(" \n Final kvar = ");disp(%i*kvar_final);
+printf(" \n Final kVA = " );disp(kVA_final);
+printf(" \n Final kVA = %f <%.2f kVA \n ",kVA_final_m,kVA_final_a);
+
+printf(" \n Final PF = %.3f lagging \n ", PF_final );
+
+printf(" \n __________________________________________________________________________");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n __________________________________________________________________________");
+printf(" \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag",kW ,kvar_original ,kVA_original,PF);
+printf(" \n Removed : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag",kW_IM,kvar_IM,kVA_IM,PF_IM);
+printf(" \n Added : \t+%.f \t\t 0 \t%.1f \t\t 1.0 ",kW_IM,kW_IM);
+printf(" \n Final : \t%d \t\tj%.f \t\t%.1f \t %.3f lag",kW ,kvar_final ,kVA_final_m,PF_final);
+printf(" \n __________________________________________________________________________");
+
diff --git a/1092/CH8/EX8.18/Example8_18.sce b/1092/CH8/EX8.18/Example8_18.sce new file mode 100755 index 000000000..f191dd590 --- /dev/null +++ b/1092/CH8/EX8.18/Example8_18.sce @@ -0,0 +1,83 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-18
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 40000 ; // Load on a factory in kW
+PF = 0.8 ; // power factor lagging of the load
+cos_theta = PF;
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+
+PF_SM = 0.8 ; // power factor leading of the synchronous motor
+cos_theta_SM = PF_SM;
+sin_theta_SM = sqrt( 1 - (cos_theta_SM)^2 );
+hp = 7500 ; // power rating of the induction motor in hp
+
+PF_IM = 0.75 ; // power factor lagging of the induction motor
+cos_theta_IM = PF_IM;
+sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );
+
+eta = 91*(1/100) ; // Efficiency of IM
+
+// Calculations
+kVA_original = kW / PF ; // Original kVA
+kvar_original = kVA_original * sin_theta ; // Original kvar
+
+
+kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW
+kVA_IM = kW_IM / PF_IM ; // Induction motor kVA
+kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar
+
+// case a
+kW_SM = ( hp * 746 ) / ( 1000 * eta ) ; // Synchronous motor kW
+kVA_SM = kW_SM / PF_SM ; // Synchronous motor kVA
+kvar_SM = kVA_SM * sin_theta_SM ; // Synchronous motor kvar
+
+kvar_final = kvar_original - kvar_IM - kvar_SM ; // final kvar
+kVA_final = kW + %i*(abs(kvar_final)); // final kVA
+kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA
+kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;
+//kVA_final_a=phase angle of kVA_final in degrees
+
+PF_final = cosd(kVA_final_a); // Final power factor
+
+// Display the result
+disp("Example 8-18 Solution : ");
+
+printf(" \n Original kVA = %d kVA \n ", kVA_original );
+printf(" \n Original kvar = \n" );disp(%i*kvar_original);
+printf(" \n a:");
+printf(" \n Synchronous motor kW = %d kW \n ", kW_SM );
+printf(" \n Synchronous motor kVA = %.f kVA \n ", kVA_SM );
+printf(" \n Synchronous motor kvar = ");disp(-%i*kvar_SM)
+
+printf(" \n Final kvar = ");disp(%i*kvar_final);
+printf(" \n Final kVA = " );disp(kVA_final);
+printf(" \n Final kVA = %f <%.2f kVA \n ",kVA_final_m,kVA_final_a);
+
+printf(" \n Final PF = %.3f lagging \n ", PF_final );
+
+printf(" \n __________________________________________________________________________");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n __________________________________________________________________________");
+printf(" \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag",kW ,kvar_original ,kVA_original,PF);
+printf(" \n Removed : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag",kW_IM,kvar_IM,kVA_IM,PF_IM);
+printf(" \n Added : \t+%.f \t\t-j%.2f \t%.1f \t\t %.1f lead",kW_SM,abs(kvar_SM),kVA_SM,PF_SM);
+printf(" \n Final : \t%d \t\tj%.2f \t%.1f \t %.3f lag",kW ,kvar_final ,kVA_final_m,PF_final);
+printf(" \n __________________________________________________________________________\n\n");
+
+printf(" \n b: ");
+printf(" \n In Ex.8-17, a 6148 kVA, unity PF, 7500 hp synchronous motor is needed.");
+printf(" \n In Ex.8-18, a 7685 kVA, 0.8 PF leading, 7500 hp synchronous motor is needed.\n");
+printf(" \n \t Ex.8-18b shows that a 0.8 PF leading,7500 hp synchronous motor ");
+printf(" \n must be physically larger than a unity PF,7500 hp synchronous motor ");
+printf(" \n because of its higher kVA rating.");
+
diff --git a/1092/CH8/EX8.19/Example8_19.sce b/1092/CH8/EX8.19/Example8_19.sce new file mode 100755 index 000000000..4284f9aad --- /dev/null +++ b/1092/CH8/EX8.19/Example8_19.sce @@ -0,0 +1,70 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-19
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA_load = 500 ; // Load of 500 kVA
+PF_load = 0.65 ; // Load operates at this PF lagging
+cos_theta_load = PF_load ;
+sin_theta_load = sqrt(1 - (cos_theta_load)^2);
+hp = 200 ; // power rating of the system in hp
+eta = 88*(1/100); // Efficiency of the system after adding the load
+PF_final = 0.85 ; // Final lagging PF after adding the load
+
+// Calculations
+kW_original = kVA_load * cos_theta_load ; // Original kW
+kvar_original = kVA_load * sin_theta_load ; // Original kvar
+
+kW_SM = ( hp * 746 ) / ( 1000 * eta ) ; // Synchronous motor kW
+
+// case a
+kW_final = kW_original + kW_SM ; // final kW of the system with the motor added
+kVA_final = kW_final / PF_final ; // final kVA of the system with the motor added
+PF_system = kW_final / kVA_final ; // Final PF of the system with the motor added
+cos_theta_system = PF_system ; // Final PF of the system with the motor added
+sin_theta_system = sqrt(1 - (cos_theta_system )^2);
+
+kvar_final = kVA_final * sin_theta_system ; // final kvar of the system with the motor added
+
+// case b
+kvar_SM = %i*kvar_final - %i*kvar_original ; // kvar rating of the sychronous motor
+
+kVA_SM = kW_SM + kvar_SM ; // kVA rating of the sychronous motor
+kVA_SM_m = abs(kVA_SM);//kVA_SM_m = magnitude of kVA_SM in kVA
+kVA_SM_a = atan(imag(kVA_SM) /real(kVA_SM))*180/%pi;
+//kVA_SM_a=phase angle of kVA_SM in degrees
+
+PF_SM = cosd(kVA_SM_a); // PF of the sychronous motor
+
+// Display the result
+disp("Example 8-19 Solution : ");
+
+printf(" \n Original kW = %.f kW \n ", kW_original );
+printf(" \n Original kvar = %.f kvar\n",kvar_original );
+printf(" \n Synchronous motor kW = %.1f kW \n ", kW_SM );
+
+printf(" \n a: Final kW = %.1f kW",kW_final);
+printf(" \n Final kVA of the system = %.f kVA",kVA_final);
+printf(" \n System PF = %.2f lagging",PF_system);
+printf(" \n Final kvar of the system = j%d (lagging)kvar\n\n",kvar_final);
+
+printf(" \n b: Synchronous motor kvar = -%.2fj(leading)kvar\n",abs(kvar_SM));
+printf(" \n Synchronous motor kVA = " );disp(kVA_SM);
+printf(" \n Synchronous motor kVA = %.f <%.1f kVA \n ", kVA_SM_m , kVA_SM_a );
+printf(" \n Synchronous motor PF = cos(%.1f) = %.3f leading \n ",kVA_SM_a,PF_SM );
+
+printf(" \n ________________________________________________");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t ±jQ \t S* ");
+printf(" \n \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ");
+printf(" \n ________________________________________________");
+printf(" \n Original : \t %d \t +j%.f %.1d \t %.2f lag",kW_original,kvar_original,kVA_load,PF_load);
+printf(" \n Added : \t %.1f \t -%.1fj %.f \t %.4f lead",kW_SM ,abs(kvar_SM),kVA_SM_m,PF_SM);
+printf(" \n Final : \t %.1f \t +j%.f %.f %.2f lag",kW_final,kvar_final,kVA_final,PF_final);
+printf(" \n ________________________________________________");
diff --git a/1092/CH8/EX8.2/Example8_2.sce b/1092/CH8/EX8.2/Example8_2.sce new file mode 100755 index 000000000..19fcee1bc --- /dev/null +++ b/1092/CH8/EX8.2/Example8_2.sce @@ -0,0 +1,74 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3- phase Y-connected synchronous motor
+P = 20 ; // No. of poles
+hp = 40 ; // power rating of the synchronous motor in hp
+V_L = 660 ; // Line voltage in volt
+beta = 5 ; // At no-load, the rotor is retarded 0.5 mechanical degree from
+// its synchronous position.
+X_s = 10 ; // Synchronous reactance in ohm
+R_a = 1.0 ; // Effective armature resistance in ohm
+
+// Calculations
+// case a
+funcprot(0); // To avoid this message "Warning : redefining function: beta"
+alpha = P * (beta/2); // The rotor shift from the synchronous position in
+// electrical degrees.
+
+// case b
+V_p = V_L / sqrt(3); // Phase voltage in volt
+E_gp = V_p ; // Generated voltage/phase at no-load in volt (given)
+E_r = (V_p - E_gp*cosd(alpha)) + %i*(E_gp*sind(alpha));
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case c
+Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm
+Z_s_m = abs(Z_s);//Z_s_m=magnitude of Z_s in ohm
+Z_s_a = atan(imag(Z_s) /real(Z_s))*180/%pi;//Z_s_a=phase angle of Z_s in degrees
+
+I_a = E_r / Z_s ; // Armature current/phase in A/phase
+I_a_m = abs(I_a);//I_a_m=magnitude of I_a in A
+I_a_a = atan(imag(I_a) /real(I_a))*180/%pi;//I_a_a=phase angle of I_a in degrees
+
+// case d
+theta = I_a_a ; // Phase angle between V_p and I_a in degrees
+P_p = V_p * I_a_m * cosd(theta); // Power per phase drawn by the motor from the bus
+P_t = 3*P_p ; // Total power drawn by the motor from the bus
+
+// csae e
+P_a = 3 * (I_a_m)^2 * R_a ; // Armature power loss at no-load in W
+P_d = (P_t - P_a)/746 ; // Internal developed horsepower at no-load
+
+// Display the results
+disp("Example 8-2 Solution : ");
+printf(" \n a: alpha = %d degrees (electrical degrees)\n",alpha );
+
+printf(" \n b: E_gp = %d V also, as given ",E_gp);
+printf(" \n E_r in V/phase = ");disp(E_r);
+printf(" \n E_r = %d <%.1f V/phase \n",E_r_m, E_r_a );
+
+printf(" \n c: Z_s in ohm/phase = ");disp(Z_s);
+printf(" \n Z_s = %.2f <%.1f ohm/phase \n",Z_s_m, Z_s_a );
+printf(" \n I_a in A/phase = ");disp(I_a);
+printf(" \n I_a = %.2f <%.2f A/phase \n ",I_a_m, I_a_a);
+
+printf(" \n d: P_p = %.2f W/phase ",P_p );
+printf(" \n P_t = %.2f W ",P_t);
+printf(" \n Note: Slight variations in power values is due to slight variations");
+printf(" \n in V_p , I_a and theta values from those of the textbook\n");
+
+
+printf(" \n e: P_a = %.f W ",P_a );
+printf(" \n P_d = %.1f hp ", P_d );
+
+
diff --git a/1092/CH8/EX8.20/Example8_20.sce b/1092/CH8/EX8.20/Example8_20.sce new file mode 100755 index 000000000..1d5154286 --- /dev/null +++ b/1092/CH8/EX8.20/Example8_20.sce @@ -0,0 +1,56 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-20
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+f_a = 400 ; // Frequency of the alternator in Hz
+f_m = 60 ; // Frequency of the motor in Hz
+
+// Calculations
+Pole_ratio = f_a / f_m ; // Ratio of no. of poles in alternator to that of motor
+// Subscript 1 below indicates 1st combination
+P_a1 = 40 ; // first combination must have 40 poles on the alternator
+P_m1 = 6 ; // first combination must have 6 poles on the synchronous motor at a speed
+S_m1 = (120*f_m) / P_m1 ; // Speed of the motor in rpm
+
+// Subscript 2 below indicates 2nd combination
+P_a2 = 80 ; // second combination must have 40 poles on the alternator
+P_m2 = 12 ; // second combination must have 12 poles on the synchronous motor at a speed
+S_m2 = (120*f_m) / P_m2 ; // Speed of the motor in rpm
+
+// Subscript 13below indicates 3rd combination
+P_a3 = 120 ; // third combination must have 40 poles on the alternator
+P_m3 = 18 ; // third combination must have 18 poles on the synchronous motor at a speed
+S_m3 = (120*f_m) / P_m3 ; // Speed of the motor in rpm
+
+// Display the result
+disp("Example 8-20 Solution : ");
+
+printf(" \n Since P_a/P_m = f_a/f_m = %d/%d, or %d/%d, the ratio of",f_a,f_m,f_a/20,f_m/20);
+printf(" \n f_a/f_m determines the combinations of poles and speed.\n");
+printf(" \n Only even multiples of the above ratio are possible,since poles ");
+printf(" \n are always in pairs, hence first three combinations are as follows \n");
+
+printf(" \n The first combination must have %d poles on the alternator and ",P_a1);
+printf(" \n %d poles on the sychronous motor at a speed = %d rpm.\n",P_m1,S_m1);
+
+printf(" \n The second combination must have %d poles on the alternator and ",P_a2);
+printf(" \n %d poles on the sychronous motor at a speed = %d rpm.\n",P_m2,S_m2);
+
+printf(" \n The third combination must have %d poles on the alternator and ",P_a3);
+printf(" \n %d poles on the sychronous motor at a speed = %d rpm.\n",P_m3,S_m3);
+
+printf(" \n ___________________________________________________________________");
+printf(" \n Combination \t Alternator Poles \t Motor Poles \t Speed (rpm)");
+printf(" \n \t P_a \t P_m \t S ");
+printf(" \n ___________________________________________________________________");
+printf(" \n First \t\t:\t %d\t\t %d \t %d",P_a1,P_m1,S_m1);
+printf(" \n Second\t\t:\t %d\t\t %d \t %d",P_a2,P_m2,S_m2);
+printf(" \n Third \t\t:\t %d\t\t %d \t %d",P_a3,P_m3,S_m3);
+printf(" \n ___________________________________________________________________");
diff --git a/1092/CH8/EX8.3/Example8_3.sce b/1092/CH8/EX8.3/Example8_3.sce new file mode 100755 index 000000000..a72a2afc2 --- /dev/null +++ b/1092/CH8/EX8.3/Example8_3.sce @@ -0,0 +1,110 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3- phase Y-connected synchronous motor
+P = 6 ; // No. of poles
+hp = 50 ; // power rating of the synchronous motor in hp
+V_L = 440 ; // Line voltage in volt
+X_s = 2.4 ; // Synchronous reactance in ohm
+R_a = 0.1 ; // Effective armature resistance in ohm
+alpha = 20 ; // The rotor shift from the synchronous position in
+// electrical degrees.
+E_gp_a = 240 ; // Generated voltage/phase in volt when the motor is under-excited(case a)
+E_gp_b = 265 ; // Generated voltage/phase in volt when the motor is under-excited(case b)
+E_gp_c = 290 ; // Generated voltage/phase in volt when the motor is under-excited(case c)
+
+// Calculations
+V_p = V_L / sqrt(3); // Phase voltage in volt
+// case a
+E_ra = (V_p - E_gp_a * cosd(alpha)) + %i*(E_gp_a * sind(alpha));
+E_ra_m = abs(E_ra);//E_ra_m=magnitude of E_ra in volt
+E_ra_a = atan(imag(E_ra) /real(E_ra))*180/%pi;//E_ra_a=phase angle of E_ra in degrees
+
+Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm
+
+I_ap1 = E_ra / Z_s ; // Armature current/phase in A/phase
+I_ap1_m = abs(I_ap1);//I_ap1_m=magnitude of I_ap1 in A
+I_ap1_a = atan(imag(I_ap1) /real(I_ap1))*180/%pi;//I_ap1_a=phase angle of I_ap1 in degrees
+
+cos_theta_a = cosd(I_ap1_a); // Power factor
+Ia_m1 = abs(I_ap1_m); // Absoulte value of magnitude of I_ap1
+
+P_d1 = 3 * (E_gp_a*Ia_m1) * cosd(160 - I_ap1_a); // // Internal developed power in W
+// 160 + I_ap1_a is the angle between E_gp_a and I_ap1
+Pd1 = abs(P_d1); // Consider absolute value of power in W for calculating hp
+
+Horse_power1 = Pd1 / 746 ; // Horsepower developed by the armature in hp
+
+// case b
+E_rb = (V_p - E_gp_b * cosd(alpha)) + %i*(E_gp_b * sind(alpha));
+E_rb_m = abs(E_rb);//E_rb_m=magnitude of E_rb in volt
+E_rb_a = atan(imag(E_rb) /real(E_rb))*180/%pi;//E_rb_a=phase angle of E_rb in degrees
+
+I_ap2 = E_rb / Z_s ; // Armature current/phase in A/phase
+I_ap2_m = abs(I_ap2);//I_ap2_m=magnitude of I_ap2 in A
+I_ap2_a = atan(imag(I_ap2) /real(I_ap2))*180/%pi;//I_ap2_a=phase angle of I_ap2 in degrees
+
+cos_theta_b = cosd(I_ap2_a); // Power factor
+Ia_m2 = abs(I_ap2_m); // Absoulte value of magnitude of I_ap2
+
+P_d2 = 3 * (E_gp_b*Ia_m2) * cosd(160 - I_ap2_a); // // Internal developed power in W
+// 160 + I_ap2_a is the angle between E_gp_b and I_ap2
+Pd2 = abs(P_d2); // Consider absolute value of power in W for calculating hp
+
+Horse_power2 = Pd2 / 746 ; // Horsepower developed by the armature in hp
+
+// case c
+E_rc = (V_p - E_gp_c * cosd(alpha)) + %i*(E_gp_c * sind(alpha));
+E_rc_m = abs(E_rc);//E_rc_m=magnitude of E_rc in volt
+E_rc_a = atan(imag(E_rc) /real(E_rc))*180/%pi;//E_rc_a=phase angle of E_rc in degrees
+
+I_ap3 = E_rc / Z_s ; // Armature current/phase in A/phase
+I_ap3_m = abs(I_ap3);//I_ap3_m=magnitude of I_ap3 in A
+I_ap3_a = atan(imag(I_ap3) /real(I_ap3))*180/%pi;//I_ap3_a=phase angle of I_ap3 in degrees
+
+cos_theta_c = cosd(I_ap3_a); // Power factor
+Ia_m3 = abs(I_ap3_m); // Absoulte value of magnitude of I_ap3
+
+P_d3 = 3 * (E_gp_c*Ia_m3) * cosd(160 - I_ap3_a); // // Internal developed power in W
+// 160 + I_ap3_a is the angle between E_gp_c and I_ap3
+Pd3 = abs(P_d3); // Consider absolute value of power in W for calculating hp
+
+Horse_power3 = Pd3 / 746 ; // Horsepower developed by the armature in hp
+
+// Display the results
+disp("Example 8-3 Solution : ");
+disp("Slight variations in power values are because of non-approximation of I_a & cos(E_gp,I_a) values during power calculations in scilab ")
+printf(" \n a: V_p = %.f <0 V \n ",V_p);
+printf(" \n E_r in V = ");disp(E_ra);
+printf(" \n E_r = %.2f <%.2f V \n ",E_ra_m,E_ra_a);
+printf(" \n I_ap in A = ");disp(I_ap1);
+printf(" \n I_ap = %.2f <%.2f A \n", I_ap1_m , I_ap1_a );
+printf(" \n cos(theta) = %.4f lagging \n ", cos_theta_a );
+printf(" \n P_d = %d W drawn from bus(motor operation)\n", P_d1 );
+printf(" \n Horsepower = %.1f hp \n\n", Horse_power1 );
+
+printf(" \n b: E_r in V = ");disp(E_rb);
+printf(" \n E_r = %.2f <%.2f V \n ",E_rb_m,E_rb_a);
+printf(" \n I_ap in A = ");disp(I_ap2);
+printf(" \n I_ap = %.2f <%.2f A \n", I_ap2_m , I_ap2_a );
+printf(" \n cos(theta) = %.4f = %.f(unity PF) \n ", cos_theta_b, cos_theta_b );
+printf(" \n P_d = %d W drawn from bus(motor operation)\n", P_d2 );
+printf(" \n Horsepower = %.1f hp \n\n", Horse_power2 );
+
+printf(" \n c: E_r in V = ");disp(E_rc);
+printf(" \n E_r = %.2f <%.2f V \n ",E_rc_m,E_rc_a);
+printf(" \n I_ap in A = ");disp(I_ap3);
+printf(" \n I_ap = %.2f <%.2f A \n", I_ap3_m , I_ap3_a );
+printf(" \n cos(theta) = %.4f leading \n ", cos_theta_c );
+printf(" \n P_d = %d W drawn from bus(motor operation)\n", P_d3 );
+printf(" \n Horsepower = %.1f hp \n\n", Horse_power3 );
+
+
diff --git a/1092/CH8/EX8.4/Example8_4.sce b/1092/CH8/EX8.4/Example8_4.sce new file mode 100755 index 000000000..d3af817aa --- /dev/null +++ b/1092/CH8/EX8.4/Example8_4.sce @@ -0,0 +1,90 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Y-connected synchronous dynamo
+P = 2 ; // No. of poles
+hp = 1000 ; // power rating of the synchronous motor in hp
+V_L = 6000 ; // Line voltage in volt
+f = 60 ; // Frequency in Hz
+R_a = 0.52 ; // Effective armature resistance in ohm
+X_s = 4.2 ; // Synchronous reactance in ohm
+P_t = 811 ; // Input power in kW
+PF = 0.8 ; // Power factor leading
+
+// Calculations
+V_p = V_L / sqrt(3); // Phase voltage in volt
+
+// case a
+cos_theta = PF ; // Power factor leading
+I_L = (P_t*1000) / ( sqrt(3) * V_L * cos_theta); // Line armature current in A
+I_ap = I_L ; // Phase armature current in A
+
+// case b
+Z_p = R_a + %i * X_s ; // Impedance per phase in ohm
+Z_p_m = abs(Z_p);//Z_p_m=magnitude of Z_p in ohm
+Z_p_a = atan(imag(Z_p) /real(Z_p))*180/%pi;//Z_p_a=phase angle of Z_p in degrees
+
+// case c
+Ia_Zp = I_L * Z_p_m ;
+E_r = Ia_Zp ;
+
+// case d
+theta = acosd(0.8); // Power factor angle in degrees
+
+// case e
+funcprot(0); // Use to avoid this message "Warning : redefining function: beta" .
+beta = Z_p_a ; //
+deba = beta + theta // Difference angle at 0.8 leading PF in degrees
+
+// case f
+// Generated voltage/phase in volt
+E_gp_f = sqrt( (E_r)^2 + (V_p)^2 - 2*E_r*V_p*cosd(deba) );
+
+// case g
+// Generated voltage/phase in volt
+E_gp_g = ( V_p + Ia_Zp * cosd(180-deba) ) + %i * ( Ia_Zp * sind(180-deba) );
+E_gp_g_m = abs(E_gp_g);//E_gp_g_m=magnitude of E_gp_g in volt
+E_gp_g_a = atan(imag(E_gp_g) /real(E_gp_g))*180/%pi;//E_gp_g_a=phase angle of E_gp_g in degrees
+
+// case h
+IaZp = Ia_Zp * expm(%i * Z_p_a * (%pi/180) ); // voltage generated by alternator 1 in volt
+IaZp_m = abs(IaZp);//IaZp_m=magnitude of IaZp in A
+IaZp_a = atan(imag(IaZp) /real(IaZp))*180/%pi;//IaZp_a=phase angle of IaZp in degrees
+IaRa = IaZp_m*cosd(IaZp_a); // Real part of IaZp
+IaXs = IaZp_m*sind(IaZp_a); // Imaginery part of IaZp
+
+cos_theta = PF ; //
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+// Generated voltage/phase in volt
+E_gp_h = ( V_p * cos_theta - IaRa ) + %i * ( V_p * sin_theta + IaXs);
+E_gp_h_m = abs(E_gp_h);//E_gp_h_m=magnitude of E_gp_h in volt
+E_gp_h_a = atan(imag(E_gp_h) /real(E_gp_h))*180/%pi;//E_gp_h_a=phase angle of E_gp_h in degrees
+
+// Display the results
+disp("Example 8-4 Solution : ");
+printf(" \n a: I_L = %.2f \n I_ap = %.2f A \n", I_L, I_ap );
+
+printf(" \n b: Z_p in ohm = ");disp(Z_p);
+printf(" \n Z_p = %.3f <%.2f ohm \n ", Z_p_m , Z_p_a );
+
+printf(" \n c: IaZp = %.1f V \n E_r = %.1f V \n ",Ia_Zp , E_r );
+
+printf(" \n d: Power factor angle,\n theta = %.2f degrees leading \n ", theta );
+
+printf(" \n e: Difference angle,\n deba = %.2f degrees \n ", deba );
+
+printf(" \n f: E_gp = %.f V \n ", E_gp_f );
+
+printf(" \n g: E_gp in V = ");disp(E_gp_g );
+printf(" \n E_gp = %d <%.2f V \n",E_gp_g_m , E_gp_g_a );
+
+printf(" \n h: E_gp in V = ");disp(E_gp_h);
+printf(" \n E_gp = %.f <%.2f V",E_gp_h_m, E_gp_h_a );
diff --git a/1092/CH8/EX8.5/Example8_5.sce b/1092/CH8/EX8.5/Example8_5.sce new file mode 100755 index 000000000..db33e8140 --- /dev/null +++ b/1092/CH8/EX8.5/Example8_5.sce @@ -0,0 +1,51 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Y-connected synchronous dynamo
+P = 2 ; // No. of poles
+hp = 1000 ; // power rating of the synchronous motor in hp
+V_L = 6000 ; // Line voltage in volt
+f = 60 ; // Frequency in Hz
+R_a = 0.52 ; // Effective armature resistance in ohm
+X_s = 4.2 ; // Synchronous reactance in ohm
+P_t = 811 ; // Input power in kW
+PF = 0.8 ; // Power factor leading
+
+// Calculated values
+E_gp = 3687 ; // Generated voltage/phase in volt
+V_p = V_L / sqrt(3); // Phase voltage in volt
+E_r = 412.8 ; // Resultant EMF across armature/phase in volt
+deba = 119.81 ; // Difference angle at 0.8 leading PF in degrees
+theta = 36.87 ; // Power factor angle in degrees
+IaXs = 409.7 ; // Voltage drop across synchronous reactance in volt
+IaRa = 50.74 ; // Voltage drop across armature resistance in volt
+
+// Calculations
+
+// Torque angle alpha in degrees calculated by different Eqns
+// case a
+alpha1 = acosd( ( E_gp^2 + V_p^2 - E_r^2 ) / ( 2*E_gp*V_p ) ); // Eq.8-12
+
+// case b
+alpha2 = asind( ( E_r * sind(deba) ) / ( E_gp ) ); // Eq.8-13
+
+// case c
+alpha3 = theta - atand( (V_p*sind(theta) + IaXs) / (V_p*cosd(theta) - IaRa) );// Eq.8-14
+
+// Display the results
+disp("Example 8-5 Solution : ");
+printf(" \n a: Using Eq.(8-12) \n alpha = %.2f degrees \n ", alpha1 );
+
+printf(" \n b: Using Eq.(8-13) \n alpha = %.2f degrees \n ", alpha2 );
+
+printf(" \n c: Using Eq.(8-14) \n alpha = %.2f degrees \n ", alpha3 );
+printf(" \n Slight variation in case c alpha is due to tan inverse value ");
+printf(" \n which was calulated to be 42.445604 degrees, instead of 42.44 degrees(textbook).")
diff --git a/1092/CH8/EX8.6/Example8_6.sce b/1092/CH8/EX8.6/Example8_6.sce new file mode 100755 index 000000000..5e3e4e3d1 --- /dev/null +++ b/1092/CH8/EX8.6/Example8_6.sce @@ -0,0 +1,59 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data as per Example 8-4
+// Y-connected synchronous dynamo
+P = 2 ; // No. of poles
+hp = 1000 ; // power rating of the synchronous motor in hp
+V_L = 6000 ; // Line voltage in volt
+f = 60 ; // Frequency in Hz
+R_a = 0.52 ; // Effective armature resistance in ohm
+X_s = 4.2 ; // Synchronous reactance in ohm
+P_t = 811 ; // Input power in kW
+PF = 0.8 ; // Power factor leading
+
+// Calculated values from Example 8-4
+E_gp = 3687 ; // Generated voltage/phase in volt
+
+I_a = 97.55 ; // Phase armature current in A
+
+phi = (42.45 - 0); // Phase angle between E_gp and I_a in degrees
+// where 42.45 and 0 are phase angles of E_gp and I_a in degrees respectively.
+
+// Calculations
+// case a
+P_p = E_gp * I_a * cosd(phi) / 1000; // Mechanical power developed per phase in kW
+
+P_t_a = 3 * P_p ; // Total mechanical power developed in kW
+
+// case b
+P_t_b = P_t_a / 0.746 ; // Internal power developed in hp at rated load
+
+// case c
+S = 120 * f / P ; // Speed of the motor in rpm
+T_int = ( P_t_b * 5252 ) / S ; // Internal torque developed in lb-ft
+
+// case d
+T_ext = ( hp * 5252 ) / 3600 ; // External torque developed in lb-ft
+eta = (T_ext / T_int) * 100 ; // Motor efficiency in percent
+
+// Display the results
+disp("Example 8-6 Solution : ");
+printf(" \n a: Similar to a dc motor, the mechanical power developed in the armature");
+printf(" \n is the product of the induced EMF per phase, the armature current");
+printf(" \n per phase, and the cosine of the angle between them.\n");
+printf(" \n P_p = %.3f kW \n P_t = %.1f kW \n", P_p, P_t_a );
+
+printf(" \n b: P_t = %.1f hp \n ", P_t_b );
+
+printf(" \n c: T_int = %.f lb-ft \n ", T_int );
+
+printf(" \n d: T_ext = %d lb-ft \n", T_ext );
+printf(" \n Motor Efficiency,\n eta = %.1f percent ", eta );
diff --git a/1092/CH8/EX8.7/Example8_7.sce b/1092/CH8/EX8.7/Example8_7.sce new file mode 100755 index 000000000..d89d99b8c --- /dev/null +++ b/1092/CH8/EX8.7/Example8_7.sce @@ -0,0 +1,95 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2000 ; // Total power consumed by a factory in kW from the transformer
+cos_theta = 0.6 ; // 0.6 lagging power factor at which power is consumed -
+// - from the transformer
+sin_theta = sqrt(1 - (cos_theta)^2);
+theta = -acosd(0.6); // power factor angle at which power is consumed -
+// - from the transformer in degrees
+
+V_L = 6000 ; // Primary line voltage of a transformer in volt
+
+P = 750 ; // kW expected to be delivered by the dc motor-generator
+
+hp = 1000 ; // hp rating of the motor(induction or synchronous)
+V_L_m = 6000 ; // Line voltage of a synchronous(or induction) motor in volt
+cos_theta_sm = 0.8 ; // 0.8 leading power factor of the synchronous motor
+theta_sm = acosd(0.8); // power factor angle of the synchronous motor in degrees
+
+cos_theta_im = 0.8 ; // 0.8 lagging power factor of the induction motor
+theta_im = -acosd(0.8); // power factor angle of the induction motor in degrees
+
+eta = 0.92 ; // Efficiency of each motor
+
+// Calculations
+// case a : using Induction Motor(IM)
+P_m = ( hp * 746 ) / eta ; // Induction(or synchronous) motor load in W
+I_1 = P_m / ( sqrt(3) * V_L_m * cos_theta_im ); // Lagging current drawn by IM in A
+
+I_1_prime = P_o * 1000 / ( sqrt(3) * V_L * cos_theta ); // Original lagging -
+// - factory load current in A
+
+// Total load current in A using Induction Motor :
+I_TM = I_1*(cosd(theta_im) + %i*sind(theta_im)) + I_1_prime*(cosd(theta) + %i*sind(theta)) ;
+I_TM_m = abs(I_TM);//I_TM_m = magnitude of I_TM in A
+I_TM_a = atan(imag(I_TM) /real(I_TM))*180/%pi;//I_TM_a=phase angle of I_TM in degrees
+
+PF_im = cosd(I_TM_a); // Overall PF using induction motor
+
+// case b: using synchronous motor
+I_s1 = P_m / ( sqrt(3) * V_L_m * cos_theta_sm ); // Lagging current drawn by IM in A
+
+// Total load current in A using synchronous motor :
+I_TSM = I_s1*(cosd(theta_sm) + %i*sind(theta_sm)) + I_1_prime*(cosd(theta) + %i*sind(theta)) ;
+I_TSM_m = abs(I_TSM);//I_TSM_m = magnitude of I_TSM in A
+I_TSM_a = atan(imag(I_TSM) /real(I_TSM))*180/%pi;//I_TSM_a=phase angle of I_TSM in degrees
+
+PF_sm = cosd(I_TSM_a); // Overall PF using Synchronous motor
+
+// case c
+percent_I_L = ( I_TM_m - I_TSM_m ) / I_TM_m * 100 ; // Percent reduction in -
+// - total load current in percent
+
+// Display the results
+printf("Note : case a,I1 calculated is around 97.53 A instead of 47.53 A(textbook).\n")
+printf(" Note : case b,Actual I_s1 imaginary part is around 58.52 instead of ");
+printf(" \n 52.52(textbook)so slight variation in I_TSM and percent ")
+printf(" \n reduction in total load current.\n")
+
+disp("Example 8-7 Solution : ");
+printf(" \n a: Induction(or sunchronous) motor load");
+printf(" \n P_m = %.f W ",P_m);
+printf(" \n Lagging current drawn by the IM = I1");
+printf(" \n I_1 = %.2f <-%.2f A \n",I_1,acosd(cos_theta_sm));
+printf(" \n I_1 in A = ");disp(I_1*cosd(-36.87)+%i*I_1*sind(-36.87));
+printf(" \n Original lagging factory load current = I_1_prime");
+printf(" \n I_1_prime in A = ");disp(I_1_prime*cosd(theta)+%i*I_1_prime*sind(theta));
+printf(" \n I_1_prime = %.1f <-%.2f A \n",I_1_prime,acosd(cos_theta));
+printf(" \n Total load current = motor load + factory load");
+printf(" \n I_TM = I_1 + I_1_prime\n");
+printf(" \n I_TM in A = ");disp(I_TM);
+printf(" \n I_TM = %.1f <%.1f A \n ",I_TM_m , I_TM_a );
+printf(" \n Overall system PF = %.4f lagging \n ", PF_im );
+
+printf(" \n b: Synchronous motor load\n I_s1 = %.2f <%.2f A\n",I_1,acosd(cos_theta_sm));
+printf(" \n I_s1 in A = ");disp(I_s1*cosd(36.87)+%i*I_s1*sind(36.87));
+printf(" \n Total load current : I_TSM = I_s1 + I_1_prime \n");
+printf(" \n I_TSM in A = ");disp(I_TSM);
+printf(" \n I_TSM = %.1f <%.1f A \n ",I_TSM_m , I_TSM_a );
+printf(" \n Overall system PF = %.1f lagging \n ", PF_sm );
+
+printf(" \n c: Percent reduction in total load current = %.1f percent \n",percent_I_L);
+
+printf(" \n d: PF improvement: Using the synchronous motor ( in lieu of the IM)");
+printf(" \n raises the total system PF from %.4f lagging to %.1f lagging.",PF_im,PF_sm);
+
+
diff --git a/1092/CH8/EX8.8/Example8_8.sce b/1092/CH8/EX8.8/Example8_8.sce new file mode 100755 index 000000000..c4442f8b2 --- /dev/null +++ b/1092/CH8/EX8.8/Example8_8.sce @@ -0,0 +1,41 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data from Ex.8-3a
+// 3- phase Y-connected synchronous motor
+P = 6 ; // No. of poles
+hp = 50 ; // power rating of the synchronous motor in hp
+V_L = 440 ; // Line voltage in volt
+X_s = 2.4 ; // Synchronous reactance in ohm
+R_a = 0.1 ; // Effective armature resistance in ohm
+alpha = 20 ; // The rotor shift from the synchronous position in
+// electrical degrees.
+E_gp = 240 ; // Generated voltage/phase in volt when the motor is under-excited
+f = 60 ; // Frequency in Hz
+
+// Calculated values from Example 8-3a
+V_p = 254 ; // Phase voltage in volt
+
+// Calculations
+// case a
+// Torque developed per phase Using Eq.(8-17a)
+S = 120 * f / P ; // Speed of the motor in rpm
+T_p = ( 7.04 * E_gp * V_p ) / ( S*X_s) * sind(alpha);
+
+// case b
+// Total horsepower developed using part a
+Horsepower = ( 3*T_p*S )/5252;
+
+// Display the results
+disp("Example 8-8 Solution : ");
+printf(" \n From given and calculated data of Ex.8-3a,\n");
+printf(" \n a: T_p = %.2f lb-ft \n ", T_p );
+
+printf(" \n b: Horsepower = %.1f hp ", Horsepower );
diff --git a/1092/CH8/EX8.9/Example8_9.sce b/1092/CH8/EX8.9/Example8_9.sce new file mode 100755 index 000000000..6eb6e417a --- /dev/null +++ b/1092/CH8/EX8.9/Example8_9.sce @@ -0,0 +1,64 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2000 ; // Total power consumed by a factory in kW
+cos_theta = 0.6 ; // 0.6 power factor at which power is consumed
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+V = 6000 ; // Line voltage in volt
+// Synchronous capacitor is used to raise the overall PF to unity
+P_loss_cap = 275 ; // Synchronous capacitor losses in kW
+
+// Calculations
+// case a
+S_o_conjugate = P_o / cos_theta ; // apparent complex power in kW
+jQ_o = S_o_conjugate * sin_theta ; // Original kilovars of lagging load
+
+// case b
+jQ_c = -jQ_o ; // Kilovars of correction needed to bring the PF to unity
+
+// case c
+R = P_loss_cap ; // Synchronous capacitor losses in kW
+S_c_conjugate = R - %i*( abs(jQ_c) ) ; // kVA rating of the synchronous capacitor
+S_c_conjugate_m = abs(S_c_conjugate);//S_c_conjugate_m = magnitude of S_c_conjugate in kVA
+S_c_conjugate_a = atan(imag(S_c_conjugate) /real(S_c_conjugate))*180/%pi;
+//S_c_conjugate_a=phase angle of S_c_conjugate in degrees
+PF = cosd(S_c_conjugate_a); // Power factor of the synchronous capacitor
+
+// case d
+I_o = S_o_conjugate * 1000 / V ; // Original current drawn from the mains in A
+
+
+// case e
+P_f = P_o + P_loss_cap ; // Total power in kW
+S_f = P_f ; // Total apparent power in kW
+S_f_m = abs(S_f);//S_f_m = magnitude of S_f in A
+S_f_a = atan(imag(S_f) /real(S_f))*180/%pi;//S_f_a=phase angle of S_f in degrees
+
+I_f = S_f * 1000 / V ; // Final current drawn from the mains after correction in A
+
+// Display the results
+disp("Example 8-9 Solution : ");
+printf(" \n a: S*o = %d kVA \n", S_o_conjugate );
+printf(" \n +jQo in kvar = ");disp(%i*jQ_o);
+
+printf(" \n b: -jQc in kvar = " );disp(%i*jQ_c);
+
+printf(" \n c: S*c in kVA = ");disp(S_c_conjugate);
+printf(" \n S*c = %.f <%.1f kVA \n", S_c_conjugate_m , S_c_conjugate_a );
+printf(" \n PF = %.3f leading \n",PF );
+
+printf(" \n d: I_o = %.1f A \n ",I_o );
+
+printf(" \n e: S_f in A = ");disp(S_f);
+printf(" \n S_f = %d <%d kVA \n" , S_f_m , S_f_a );
+printf(" \n I_f = %.1f A \n ", I_f);
+
+printf(" \n f: See Fig.8-25.");
diff --git a/1092/CH9/EX9.1/Example9_1.sce b/1092/CH9/EX9.1/Example9_1.sce new file mode 100755 index 000000000..87435c925 --- /dev/null +++ b/1092/CH9/EX9.1/Example9_1.sce @@ -0,0 +1,34 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+phase = 3 ; // Number of phases
+n = 3 ; // Slots per pole per phase
+f = 60 ; // Line frequency in Hz
+
+// Calculations
+// case a
+P = 2 * n ; // Number of poles produced
+Total_slots = n * P * phase ; // Total number of slots on the stator
+
+// case b
+S_b = (120*f)/P ; // Speed in rpm of the rotating magnetic field
+
+// case c
+f_c = 50 ; // Changed line frequency in Hz
+S_c = (120*f_c)/P ; // Speed in rpm of the rotating magnetic field
+
+// Display the results
+disp("Example 9-1 Solution : ");
+printf(" \n a: P = %d poles \n Total slots = %d slots \n", P ,Total_slots );
+
+printf(" \n b: S = %d rpm @ f = %d Hz \n ", S_b , f );
+
+printf(" \n c: S = %d rpm @ f = %d Hz ", S_c ,f_c );
diff --git a/1092/CH9/EX9.10/Example9_10.sce b/1092/CH9/EX9.10/Example9_10.sce new file mode 100755 index 000000000..6a5e01bd5 --- /dev/null +++ b/1092/CH9/EX9.10/Example9_10.sce @@ -0,0 +1,64 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 4 ; // Number of poles in WRIM
+f = 60 ; // Frequency in Hz
+V = 220 ; // Line voltage in volt
+V_p = 220 ; // Phase voltage in volt (delta connection)
+hp_WRIM = 1 ; // Power rating of WRIM in hp
+S_r = 1740 ; // Full-load rated speed in rpm
+R_r = 0.3 ; // rotor resistance per phase in ohm/phase
+R_x = 0.7 ; // Added resistance in ohm/phase
+X_lr =1 ; // Locked rotor reactance in ohm
+
+// Calculations
+S = (120*f)/P ; // Speed in rpm of the rotating magnetic field
+// case a
+E_lr = V_p / 4 ; // Locked-rotor voltage per phase
+
+// case b
+s = ( S - S_r)/S ; // slip
+I_r = E_lr / sqrt( (R_r/s)^2 + (X_lr)^2 ); // Rotor current per phase at rated speed
+
+// case c
+P_in = ((I_r)^2 * R_r)/s ; // Rated rotor power input per phase
+
+// case d
+P_RL = (I_r)^2 * R_r ; // Rated copper loss per phase
+
+// case e
+P_d_W = P_in - P_RL ; // Rotor power developed per phase in W
+P_d_hp = P_d_W/746 ; // Rotor power developed per phase in hp
+
+// case f
+hp = P_d_hp ; // Rotor power developed per phase in hp
+T_d1 = (hp*5252)/S_r ; // Rotor torque developed in lb-ft per phase by method 1
+T_d2 = 7.04*(P_in/S) ; // Rotor torque developed in lb-ft per phase by method 2
+
+T_dm = 3*T_d1 ; // Total rotor torque in lb-ft
+
+// Display the results
+disp("Example 9-10 Solution : ");
+printf(" \n a: Locked-rotor voltage per phase : \n E_lr = %d V \n ",E_lr);
+
+printf(" \n b: slip : \n s = %.2f \n",s);
+printf(" \n Rotor current per phase at rated speed:\n I_r = %.3f A/phase \n ",I_r);
+
+printf(" \n c: Rated rotor power input per phase :\n P_in = %d W/phase \n ",P_in);
+
+printf(" \n d: Rated copper loss per phase : \n P_RL = %.2f W \n ",P_RL);
+
+printf(" \n e: Rotor power developed per phase in W :\n P_d = %.1f W/phase ",P_d_W);
+printf(" \n\n Rotor power developed per phase in hp :\n P_d = %.2f hp/phase \n ",P_d_hp);
+
+printf(" \n f: Rotor torque developed in lb-ft per phase :\n T_d = %.1f lb-ft (method 1)",T_d1);
+printf(" \n\n T_d = %.1f lb-ft (method 2)",T_d2);
+printf(" \n\n Total rotor torque : \n T_dm = %.1f lb-ft )\n ",T_dm);
diff --git a/1092/CH9/EX9.11/Example9_11.sce b/1092/CH9/EX9.11/Example9_11.sce new file mode 100755 index 000000000..0cf623d34 --- /dev/null +++ b/1092/CH9/EX9.11/Example9_11.sce @@ -0,0 +1,81 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase WRIM
+V_L = 208 ; // Voltage rating of the WRIM in volt
+P = 6 ; // Number of poles in WRIM
+f = 60 ; // Frequency in Hz
+P_o = 7.5 ; // Power rating of WRIM in hp
+S_r = 1125 ; // Full-load rotor speed in rpm
+R_r = 0.08 ; // Rotor resistance in ohm/phase
+X_lr = 0.4 ; // Locked rotor resistance in ohm/phase
+
+// Calculations
+S = (120*f)/P ; // Speed in rpm of the rotating magnetic field
+// case a
+E_lr = (V_L / sqrt(3))/2 ; // Locked rotor voltage per phase
+
+// case b
+s = (S - S_r)/S ; // Full-load rated slip
+I_r = E_lr / sqrt( (R_r/s)^2 + (X_lr)^2 ); // Rotor current in A per phase at rated speed
+
+// case c
+P_in = ( (I_r)^2 * R_r )/s ; // Rated rotor power input per phase in (W/phase)
+
+// case d
+P_RL = ( (I_r)^2 * R_r ); // Rated rotor copper loss per phase (in W/phase)
+
+// case e
+// Subscript W in P_d indicates calculating P_d in W
+P_d_W = P_in - P_RL ; // Rotor power developed per phase (in W/phase)
+// Subscript hp in P_d indicates calculating P_d in hp
+P_d_hp = P_d_W/746 ; // Rotor power developed per phase (in hp/phase)
+
+// case f
+// subscript 1 in T_d indicates method 1 for calculating T_d
+hp = P_d_hp ;
+T_d1 = (hp*5252)/S_r ; // Rotor torque developed per phase in lb-ft
+
+// subscript 2 in T_d indicates method 2 for calculating T_d
+T_d2 = 7.04*(P_in/S); // Rotor torque developed per phase in lb-ft
+
+// case g
+T_dm = 3*T_d1 ; // Total rotor torque in lb-ft
+
+// case h
+T_o = 7.04*(P_o*746)/S_r ; // Total output rotor torque in lb-ft
+
+// Display the results
+disp("Example 9-11 Solution : ");
+
+printf(" \n Note: Slight variations in the answers I_r,P_in,P_RL,P_d,T_d ");
+printf(" \n are because of non-approximation of E_lr and (R_r/s)^2 + (X_lr)^2");
+printf(" \n while calulating in scilab.\n");
+
+printf(" \n a: Locked rotor voltage per phase :\n E_lr = %d V\n",E_lr);
+
+printf(" \n b: slip :\n s = %.4f ",s);
+printf(" \n\n Rotor current per phase at rated speed :\n I_r = %.2f A/phase\n",I_r);
+
+printf(" \n c: Rated rotor power input per phase :\n P_in = %.f W/phase\n",P_in);
+
+printf(" \n d: Rated rotor copper loss per phase :\n P_RL = %.1f W/phase\n",P_RL);
+
+printf(" \n e: Rotor power developed per phase ");
+printf(" \n P_d = %.f W/phase \n P_d = %.2f hp/phase\n",P_d_W,P_d_hp);
+
+printf(" \n f: Rotor torque developed per phase : ");
+printf(" \n (method 1)\n T_d = %.1f lb-ft/phase",T_d1);
+printf(" \n\n (method 2)\n T_d = %.1f lb-ft/phase\n",T_d2);
+
+printf(" \n g: Total rotor torque : \n T_dm = %d lb-ft\n",T_dm);
+
+printf(" \n h: Total output rotor torque : \n T_o = %d lb-ft",T_o);
diff --git a/1092/CH9/EX9.12/Example9_12.sce b/1092/CH9/EX9.12/Example9_12.sce new file mode 100755 index 000000000..34e176054 --- /dev/null +++ b/1092/CH9/EX9.12/Example9_12.sce @@ -0,0 +1,47 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-12 + +clear; clc; close; // Clear the work space and console. + +// Given data as per Ex.9-10 +P = 4 ; // Number of poles in WRIM +f = 60 ; // Frequency in Hz +V = 220 ; // Line voltage in volt +V_p = 220 ; // Phase voltage in volt (delta connection) +hp_WRIM = 1 ; // Power rating of WRIM in hp +S_r = 1740 ; // Full-load rated speed in rpm +R_r = 0.3 ; // rotor resistance per phase in ohm/phase +R_x = 0.7 ; // Added resistance in ohm/phase +X_lr = 1 ; // Locked rotor reactance in ohm + +// Calculations from Ex.9-10 +E_lr = V_p / 4 ; // Locked-rotor voltage per phase +S = (120*f)/P ; // Speed in rpm of the rotating magnetic field + +// Calculations (Ex.9-12) +P_in = (E_lr)^2 / (2*X_lr); // rotor power input(RPI) in W/phase +P_in_total = P_in * 3 ; // Total 3-phase rotor power input(RPI) in W + +T_max = 7.04*(P_in_total/S); // Maximum torque developed in lb-ft + +s_b = R_r / X_lr ; // Slip + +s = s_b; +S_r = S*(1 - s); // Rotor speed in rpm for T_max + +// Display the results +disp("Example 9-12 Solution : "); + +printf(" \n Rotor power input (RPI) per phase is : "); +printf(" \n P_in = %.1f W/phase \n",P_in); + +printf(" \n The total 3-phase rotor power input (RPI) is : "); +printf(" \n P_in = %.1f W\n",P_in_total); + +printf(" \n Substituting in Eq.(9-19),\n T_max = %.2f lb-ft\n",T_max); +printf(" \n Then, s_b = %.1f \n and S_r = %d rpm",s_b,S_r); diff --git a/1092/CH9/EX9.13/Example9_13.sce b/1092/CH9/EX9.13/Example9_13.sce new file mode 100755 index 000000000..063a6bf6b --- /dev/null +++ b/1092/CH9/EX9.13/Example9_13.sce @@ -0,0 +1,35 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-13 + +clear; clc; close; // Clear the work space and console. + +// Given data as per Ex.9-10 +P = 4 ; // Number of poles in WRIM +f = 60 ; // Frequency in Hz +V = 220 ; // Line voltage in volt +V_p = 220 ; // Phase voltage in volt (delta connection) +hp_WRIM = 1 ; // Power rating of WRIM in hp +S_r = 1740 ; // Full-load rated speed in rpm +R_r = 0.3 ; // rotor resistance per phase in ohm/phase +R_x = 0.7 ; // Added resistance in ohm/phase +X_lr = 1 ; // Locked rotor reactance in ohm + +// Calculations +E_lr = V_p / 4 ; // Locked-rotor voltage per phase +S = (120*f)/P ; // Speed in rpm of the rotating magnetic field + +// Total 3-phase rotor power input(RPI) in W +P_in = 3 * ( (E_lr)^2 ) / ( (R_r)^2 + (X_lr)^2 ) * R_r ; + +T_s = 7.04 * (P_in/S); // Starting torque developed in lb-ft + +// Display the results +disp("Example 9-13 Solution : "); + +printf(" \n P_in = %.f W \n",P_in); +printf(" \n From Eq.(9-19),starting torque is : \n T_s = %.2f lb-ft",T_s); diff --git a/1092/CH9/EX9.14/Example9_14.sce b/1092/CH9/EX9.14/Example9_14.sce new file mode 100755 index 000000000..e1715c743 --- /dev/null +++ b/1092/CH9/EX9.14/Example9_14.sce @@ -0,0 +1,29 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-14 + +clear; clc; close; // Clear the work space and console. + +// Given data +T_max = 17.75 ; // Maximum torque developed in lb-ft +s_max = 0.3 ; // Slip for which T_max occurs +s_a = 0.0333 ; // slip (case a) +s_b = 1.0 ; // slip (case b) + +// Calculations +// Subscript a in T indicates case a +T_a = T_max * ( 2 / ((s_max/s_a) + (s_a/s_max)) ); // Full-load torque in lb-ft + +// Subscript b in T indicates case b +T_b = T_max * ( 2 / ((s_max/s_b) + (s_b/s_max)) ); // Starting torque in lb-ft + +// Display the results +disp("Example 9-14 Solution : "); + +printf(" \n a: Full-load torque at slip = %.4f \n T = %.1f lb-ft\n",s_a,T_a); + +printf(" \n b: Starting torque at slip = %.1f \n T = %.2f lb-ft\n",s_b,T_b); diff --git a/1092/CH9/EX9.15/Example9_15.sce b/1092/CH9/EX9.15/Example9_15.sce new file mode 100755 index 000000000..22d24921c --- /dev/null +++ b/1092/CH9/EX9.15/Example9_15.sce @@ -0,0 +1,143 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-15 + +clear; clc; close; // Clear the work space and console. + +// Given data +// 3-phase Y-connected SCIM +P = 4 ; // Number of poles in SCIM +S_r = 1746 ; // Rated rotor speed in rpm +V = 220 ; // Voltage rating of SCIM in volt +f = 60 ; // Frequency in Hz +P_hp = 10 ; // power rating of SCIM in hp +R_a = 0.4 ; // Armature resistance in ohm +R_r = 0.14 ; // Rotor resistance in ohm +jXm = 16 ; // Reactance in ohm +jXs = 0.35 ; // Synchronous reactance in ohm +jXlr = 0.35 ; // Locked rotor reactance in ohm +P_r_total = 360 ; // Total rotational losses in W + +// Calculations +V_p = V / sqrt(3); // Voltage per phase in volt + +S = (120*f)/P ; // Speed in rpm of the rotating magnetic field +// preliminary calculations +s = ( S - S_r)/S ; // slip + +disp("Example 9-15 :"); + +printf(" \n From Fig.9-13,using the format method of mesh analysis,we may"); +printf(" \n write the array by inspection :\n"); +printf(" \n __________________________________________________________"); +printf(" \n I_1(A) \t\t I_2(A) \t\t V(volt)"); +printf(" \n __________________________________________________________"); +printf(" \n (0.4 + j16.35) \t -(0 + j16) \t\t (127 + j0)"); +printf(" \n -(0 + j16) \t\t (4.67 + j16.35) \t 0"); +printf(" \n __________________________________________________________"); + +A = [ (0.4 + %i*16.35) -%i*16 ; (-%i*16) (4.67 + %i*16.35) ]; // Matrix containing above mesh eqns array +delta = det(A); // Determinant of A + +// case a : Stator armature current I_p in A +I_p = det( [ (127+%i*0) (-%i*16) ; 0 (4.67 + %i*16.35) ] ) / delta ; +I_p_m = abs(I_p);//I_p_m=magnitude of I_p in A +I_p_a = atan(imag(I_p) /real(I_p))*180/%pi;//I_p_a=phase angle of I_p in degrees +I_1 = I_p ; // Stator armature current in A + +// case b : Rotor current I_r per phase in A +I_r = det( [ (0.4 + %i*16.35) (127+%i*0) ; (-%i*16) 0 ] ) / delta ; +I_r_m = abs(I_r);//I_r_m=magnitude of I_r in A +I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees + +// case c +theta_1 = I_p_a ; // Motor PF angle in degrees +cos_theta1 = cosd(theta_1); // Motor PF + +// case d +I_p = I_p_m ; // Stator armature current in A +SPI = V_p * I_p * cos_theta1 ; // Stator Power Input in W + +// case e +SCL = (I_p)^2 * R_a ; // Stator Copper Loss in W + +// case f +// Subscripts 1 and 2 for RPI indicates two methods of calculating RPI +RPI_1 = SPI - SCL ; // Rotor Power Input in W +RPI_2 = (I_r_m)^2 * (R_r/s); // Rotor Power Input in W +RPI =RPI_1 ; + +// case g +// Subscripts 1 , 2 and 3 for RPD indicates three methods of calculating RPD +RPD_1 = RPI * ( 1 - s ); // Rotor Power Developed in W +RCL = s*(RPI); // Rotor copper losses in W +RPD_2 = RPI - RCL ; // Rotor Power Developed in W +RPD_3 = (I_r_m)^2 * R_r * ((1-s)/s); // Rotor Power Developed in W +RPD = RPD_1 ; + +// case h +P_r = P_r_total / 3 ; // Rotational Losses per phase in W +P_o = RPD - P_r ; // Rotor power per phase in W +P_to = 3*P_o ; // Total rotor power in W + +// case i +T = 7.04 * (P_to/S_r); // Total 3-phase torque in lb-ft + +// case j +P_t = P_to ; +hp = P_t / 746 ; // Output horsepower + +// case k +P_in = SPI ; // Input power to stator in W +eta = P_o / P_in * 100 ; // Motor efficiency at rated load + +// Display the results +disp("Solution : "); +printf(" \n Preliminary calculations\n"); +printf(" \n Slip : s = %.2f \n R_r/s = %.2f ohm \n",s,R_r/s); + +printf(" \n Determinant Δ = ");disp(delta); + +printf(" \n a: Stator armature current :\n I_p in A = ");disp(I_1); +printf(" \n I_p = I_1 = %.2f <%.2f A \n ",I_p_m , I_p_a ); + +printf(" \n b: Rotor current per phase :\n I_r in A = ");disp(I_r); +printf(" \n I_r = I_2 = %.3f <%.2f A \n ",I_r_m , I_r_a ); + +printf(" \n c: Motor PF :\n cosӨ1 = %.4f \n",cos_theta1); + +printf(" \n d: Stator Power Input :\n SPI = %d W \n",SPI); + +printf(" \n e: Stator Copper Loss :\n SCL = %.f W \n",SCL); + +printf(" \n f: Rotor Power Input :\n RPI = %d W(method 1) ", RPI_1); +printf(" \n RPI = %.f W (method 2)\n",RPI_2); +printf(" \n Note: RPI calculated by 2nd method slightly varies from that of"); +printf(" \n textbook value because of non-approximation of I_r while"); +printf(" \n calculating in scilab.\n") + +printf(" \n g: Rotor Power Developed :\n RPD = %.f W \n",RPD_1); +printf(" \n Rotor copper loss :\n RCL = %d W\n",RCL); +printf(" \n RPD = %.f W \n RPD = %d W \n ",RPD_2,RPD_3); + +printf(" \n h: Rotor power per phase :\n P_o/φ = %f W/φ ",P_o); +printf(" \n\n Total rotor power:\n P_to = %f W \n",P_to); +printf(" \n Above P_o/φ and P_to values are not approximated while calculating in "); +printf(" \n SCILAB.So,they vary slightly from textbook values.\n"); + +printf(" \n i: Total 3-phase output torque :\n T = %.f lb-ft\n",T); + +printf(" \n j: Output horsepower : \n hp = %.1f hp \n",hp); + +printf(" \n k: Motor efficiency at rated load :\n η = %.1f percent \n",eta) + +printf(" \n Power flow diagram (per phase)\n"); +printf(" \n SPI----------> RPI---------> RPD----------> P_o"); +printf(" \n (%d W) | (%d W) | (%d W) | (%d W)",SPI,RPI_1,RPD_3,P_o); +printf(" \n | | |"); +printf(" \n SCL RCL P_r"); +printf(" \n (%.f W) (%d W) (%d W)",SCL,RCL,P_r); diff --git a/1092/CH9/EX9.16/Example9_16.sce b/1092/CH9/EX9.16/Example9_16.sce new file mode 100755 index 000000000..ddd6436ec --- /dev/null +++ b/1092/CH9/EX9.16/Example9_16.sce @@ -0,0 +1,52 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-16 + +clear; clc; close; // Clear the work space and console. + +// Given data +// three-phase SCIM +V = 208 ; // Rated voltage in volt +P_o = 15 ; // Rated power in hp +I = 42 ; // Rated current in A +I_st = 252 ; // Starting current in A +T_st = 120 ; // Full-voltage starting torque in lb-ft +tap = 60*(1/100) ; // Tapping in % employed by compensator + +// Calculations +// case a +I_sm = tap * I_st ; // Motor starting current in A at reduced voltage + +// case b +I_L = tap * I_sm ; // Motor line current in A(neglecting tarnsformer exciting +// current and losses) + +// case c +T_s = (tap)^2 * T_st ; // Motor starting torque at reduced voltage in lb-ft + +// case d +percent_I_L = I_L / I_st * 100 ; // Percent line current at starting + +// case e +percent_T_st = T_s / T_st * 100 ; // Percent motor starting torque + +// Display the results +disp("Example 9-16 Solution : "); + +printf(" \n a: Motor starting current at reduced voltage : "); +printf(" \n I_sm = %.1f A to the motor.\n",I_sm); + +printf(" \n b: Motor line current neglecting tarnsformer exciting current and losses :"); +printf(" \n I_L = %.2f A drawn from the mains.\n",I_L); + +printf(" \n c: Motor starting torque at reduced voltage :\n T_s = %.1f lb-ft\n",T_s); + +printf(" \n d: Percent line current at starting : "); +printf(" \n = %.f percent of line current at full voltage.\n",percent_I_L); + +printf(" \n e: Percent motor starting torque : "); +printf(" \n = %d percent of starting torque at full voltage.\n",percent_T_st); diff --git a/1092/CH9/EX9.17/Example9_17.sce b/1092/CH9/EX9.17/Example9_17.sce new file mode 100755 index 000000000..d0fb0539f --- /dev/null +++ b/1092/CH9/EX9.17/Example9_17.sce @@ -0,0 +1,77 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-17 + +clear; clc; close; // Clear the work space and console. + +// Given data +// three-phase SCIM +V_o = 220 ; // Rated voltage in volt +P = 4 ; // Number of poles in SCIM +P_o = 10 ; // Rated power in hp +f = 60 ; // Frequency in Hz(assume,not given) +T_o = 30 ; // Rated torque in lb-ft +S_r = 1710 ; // Rated rotor speed in rpm +V_n1 = 242 ; // Impressed stator voltage in volt(case a) +V_n2 = 198 ; // Impressed stator voltage in volt(case b) + +// Calculations +S = (120*f)/P ; // Speed in rpm of the rotating magnetic field +// case a : Impressed stator voltage = 242 V +s_o = (S - S_r)/S ; // Rated slip + +T_n1 = T_o * (V_n1/V_o)^2 ; // New torque in lb-ft + +s_n1 = s_o * (T_o/T_n1); // New slip + +S_rn1 = S*(1 - s_n1); + +// case b : Impressed stator voltage = 198 V +T_n2 = T_o * (V_n2/V_o)^2 ; // New torque in lb-ft + +s_n2 = s_o * (T_o/T_n2); // New slip + +S_rn2 = S*(1 - s_n2); + +// case c +// Subscript a in percent_slip and percent_speed indicates part a +percent_slip_a = (s_o - s_n1)/s_o * 100 ; // Percent change in slip in part(a) + +percent_speed_a = (S_rn1 - S_r)/S_r * 100; // Percent change in speed in part(a) + +// case d +// Subscript b in percent_slip and percent_speed indicates part b +percent_slip_b = (s_n2 - s_o)/s_o * 100 ; // Percent change in slip in part(b) + +percent_speed_b = (S_r - S_rn2)/S_r * 100; // Percent change in speed in part(b) + +// Display the results +disp("Example 9-17 Solution : "); + +printf(" \n a: Rated slip :\n s = %.2f\n",s_o); +printf(" \n For impressed stator voltage = %d V \n ",V_n1); +printf(" \n New torque :\n T_n = %.1f lb-ft \n ",T_n1); +printf(" \n New slip :\n s_n = %f \n ",s_n1); +printf(" \n New rotor speed :\n S_r = %f rpm \n",S_rn1); + +printf(" \n b: For impressed stator voltage = %d V \n ",V_n2); +printf(" \n New torque :\n T_n = %.1f lb-ft \n ",T_n2); +printf(" \n New slip :\n s_n = %f \n ",s_n2); +printf(" \n New rotor speed :\n S_r = %f rpm \n",S_rn2); + +printf(" \n c: Percent change in slip in part(a)"); +printf(" \n = %.1f percent decrease.\n",percent_slip_a); +printf(" \n Percent change in speed in part(a)"); +printf(" \n = %.2f percent increase \n",percent_speed_a); + +printf(" \n d: Percent change in slip in part(b)"); +printf(" \n = %.2f percent increase.\n",percent_slip_b); +printf(" \n Percent change in speed in part(b)"); +printf(" \n = %.2f percent decrease\n",percent_speed_b); + +printf(" \n SLIGHT VARIATIONS IN PERCENT CHANGE IN SLIP AND SPEED ARE DUE TO"); +printf(" \n NON-APPROXIMATION OF NEW SLIPS AND NEW SPEEDS CALCULATED IN SCILAB.") diff --git a/1092/CH9/EX9.18/Example9_18.sce b/1092/CH9/EX9.18/Example9_18.sce new file mode 100755 index 000000000..8a77f567b --- /dev/null +++ b/1092/CH9/EX9.18/Example9_18.sce @@ -0,0 +1,70 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-18 + +clear; clc; close; // Clear the work space and console. + +// Given data +// three-phase WRIM +V_o = 220 ; // Rated voltage in volt +P_o = 10 ; // Rated power in hp +P = 4 ; // Number of poles in WRIM(assumption) +f = 60 ; // Frequency in Hz(assume,not given) +R_ro = 0.3 ; // Rotor resistance in ohm +T_o = 30 ; // Rated torque in lb-ft +S_r = 1750 ; // Rated rotor speed in rpm +R_r_ext = 1.7 ; // External rotor resistance in ohm/phase inserted in the rotor ckt +R_rn = R_ro + R_r_ext ; // Total rotor resistance in ohm + +V_n1 = 240 ; // Impressed stator voltage in volt(case a) +V_n2 = 208 ; // Impressed stator voltage in volt(case b) +V_n3 = 110 ; // Impressed stator voltage in volt(case c) + +// Calculations +S = (120*f)/P ; // Speed in rpm of the rotating magnetic field + +// case a : Impressed stator voltage = 240 V +s_o = (S - S_r)/S ; // Rated slip + +T_n1 = T_o * (V_n1/V_o)^2 ; // New torque in lb-ft + +s_n1 = s_o * (T_o/T_n1) * (R_rn/R_ro); // New slip + +S_rn1 = S*(1 - s_n1); + +// case b : Impressed stator voltage = 208 V +T_n2 = T_o * (V_n2/V_o)^2 ; // New torque in lb-ft + +s_n2 = s_o * (T_o/T_n2) * (R_rn/R_ro); // New slip + +S_rn2 = S*(1 - s_n2); + +// case c : Impressed stator voltage = 110 V +T_n3 = T_o * (V_n3/V_o)^2 ; // New torque in lb-ft + +s_n3 = s_o * (T_o/T_n3) * (R_rn/R_ro); // New slip + +S_rn3 = S*(1 - s_n3); + +// Display the results +disp("Example 9-18 Solution : "); + +printf(" \n a: Rated slip :\n s = %f\n",s_o); +printf(" \n For impressed stator voltage = %d V \n ",V_n1); +printf(" \n New torque :\n T_n = %.1f lb-ft \n ",T_n1); +printf(" \n New slip :\n s_n = %f \n ",s_n1); +printf(" \n New rotor speed :\n S_r = %f rpm \n",S_rn1); + +printf(" \n b: For impressed stator voltage = %d V \n ",V_n2); +printf(" \n New torque :\n T_n = %.2f lb-ft \n ",T_n2); +printf(" \n New slip :\n s_n = %f \n ",s_n2); +printf(" \n New rotor speed :\n S_r = %f rpm \n",S_rn2); + +printf(" \n c: For impressed stator voltage = %d V \n ",V_n3); +printf(" \n New torque :\n T_n = %.1f lb-ft \n ",T_n3); +printf(" \n New slip :\n s_n = %f \n ",s_n3); +printf(" \n New rotor speed :\n S_r = %f rpm \n",S_rn3); diff --git a/1092/CH9/EX9.19/Example9_19.sce b/1092/CH9/EX9.19/Example9_19.sce new file mode 100755 index 000000000..b6d9bda71 --- /dev/null +++ b/1092/CH9/EX9.19/Example9_19.sce @@ -0,0 +1,53 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS +// Example 9-19 + +clear; clc; close; // Clear the work space and console. + +// Given data +P = 8 ; // Number of poles in WRIM +f = 60 ; // Operating frequency of the WRIM in Hz +/// WRIM is driven by variable-speed prime mover as a frequency changer +S_con_a1 = 1800 ; // Speed of the convertor in rpm +S_con_a2 = 450 ; // Speed of the convertor in rpm + +f_con_b1 = 25 ; // Frequency of an induction converter in Hz +f_con_b2 = 400 ; // Frequency of an induction converter in Hz +f_con_b3 = 120 ; // Frequency of an induction converter in Hz + +// Calculations +S = (120*f)/P ; // Speed in rpm of the rotating magnetic field + +// case a +// Subscript a1 in f_con indicates case a 1st frequecy in Hz +f_con_a1 = f*(1 + S_con_a1/S); // Frequency of an induction converter in Hz + +// Subscript a2 in f_con indicates case a 2nd frequency in Hz +f_con_a2 = f*(1 - S_con_a2/S); // Frequency of an induction converter in Hz + +// case b +// Subscript b1 in S-con indicates case b 1st speed of converter in rpm +S_con_b1 = ( -1 + f_con_b1/f) * S ; // Speed of the convertor in rpm + +// Subscript b2 in S-con indicates case b 2nd speed of converter in rpm +S_con_b2 = ( -1 + f_con_b2/f) * S ; // Speed of the convertor in rpm + +// Subscript b3 in S-con indicates case b 3rd speed of converter in rpm +S_con_b3 = ( -1 + f_con_b3/f) * S ; // Speed of the convertor in rpm + + +// Display the results +disp("Example 9-19 Solution : "); + +printf(" \n Using Eq.(9-26),\n"); + +printf(" \n a: f_con = %d Hz for %d rpm in opposite direction\n",f_con_a1,S_con_a1); +printf(" \n f_con = %d Hz for %d rpm in same direction\n",f_con_a2,S_con_a2); + +printf(" \n b: 1. S_con = %.f rpm, or %.f rpm in same direction.\n",S_con_b1,abs(S_con_b1)); +printf(" \n 2. S_con = %d rpm in opposite direction.\n",S_con_b2); +printf(" \n 3. S_con = %d rpm in opposite direction to rotating stator flux.\n",S_con_b3); diff --git a/1092/CH9/EX9.2/Example9_2.sce b/1092/CH9/EX9.2/Example9_2.sce new file mode 100755 index 000000000..28f36e6da --- /dev/null +++ b/1092/CH9/EX9.2/Example9_2.sce @@ -0,0 +1,35 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+
+s_a = 5*(1/100); // Slip (case a)
+s_b = 7*(1/100); // Slip (case b)
+
+// Given data and calculated values from Ex.9-1
+f_a = 60 ; // Line frequency in Hz (case a)
+f_b = 50 ; // Line frequency in Hz (case b)
+S_a = 1200 ; // Speed in rpm of the rotating magnetic field (case a)
+S_b = 1000 ; // Speed in rpm of the rotating magnetic field (case b)
+
+// Calculations
+
+// case a
+S_r_a = S_a * ( 1 - s_a ); // Rotor speed in rpm when slip is 5% (case a)
+
+// case b
+S_r_b = S_b * ( 1 - s_b ); // Rotor speed in rpm when slip is 7% (case b)
+
+// Display the results
+disp("Example 9-2 Solution : ");
+
+printf(" \n a: S_r = %.f rpm @ s = %.2f \n ", S_r_a ,s_a );
+
+printf(" \n b: S_r = %.f rpm @ s = %.2f ", S_r_b ,s_b );
diff --git a/1092/CH9/EX9.3/Example9_3.sce b/1092/CH9/EX9.3/Example9_3.sce new file mode 100755 index 000000000..5a5029e16 --- /dev/null +++ b/1092/CH9/EX9.3/Example9_3.sce @@ -0,0 +1,37 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 4 ; // Number of poles in Induction motor
+f = 60 ; // Frequency in Hz
+s_f = 5*(1/100) ; // Full-load rotor slip
+
+// Calculations
+
+// case a
+// slip, s = (S -S_r)/S ;
+// where S = Speed in rpm of the rotating magnetic field and
+// S_r = Speed in rpm of the rotor
+s = 1 ; // Slip = 1, at the instant of starting, since S_r is zero
+f_r_a = s * f ; // Rotor frequency in Hz at the instant of starting
+
+// case b
+f_r_b = s_f * f ;// Full-load rotor frequency in Hz
+
+// Display the results
+disp("Example 9-3 Solution : ");
+
+printf(" \n a: At the instant of starting, slip s = (S -S_r)/S ; ");
+printf(" \n where S_r is the rotor speed. Since the rotor speed at the ");
+printf(" \n instant of starting is zero, s = (S - 0)/S = 1 , or unity slip.");
+printf(" \n\n The rotor frequency is \n f_r = %d Hz \n\n ", f_r_a);
+
+printf(" \n b: At full-load,the slip is 5 percent(as given), and therefore");
+printf(" \n s = %.2f \n f_r = %d Hz " , s_f , f_r_b);
diff --git a/1092/CH9/EX9.4/Example9_4.sce b/1092/CH9/EX9.4/Example9_4.sce new file mode 100755 index 000000000..d6b0a3377 --- /dev/null +++ b/1092/CH9/EX9.4/Example9_4.sce @@ -0,0 +1,30 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 4 ; // Number of poles in the IM
+hp = 50 ; // rating of the IM in hp
+V_o = 208 ; // Voltage rating of the IM in volt
+T_orig = 225 ; // Starting torque in lb-ft
+I_orig = 700 ; // Instantaneous startign current in A at rated voltage
+V_s = 120 ; // Reduced 3-phase voltage supplied in volt
+
+// Calculations
+// case a
+T_s = T_orig * (V_s/V_o)^2 ; // Starting torque in lb-ft after application of V_s
+
+// case b
+I_s = I_orig * (V_s/V_o) ; // Starting current in A after application of V_s
+
+// Display the results
+disp("Example 9-4 Solution : ");
+printf(" \n a: Starting torque :\n T_s = %.f lb-ft \n",T_s );
+
+printf(" \n b: Starting current :\n I_s = %d A \n",I_s );
diff --git a/1092/CH9/EX9.5/Example9_5.sce b/1092/CH9/EX9.5/Example9_5.sce new file mode 100755 index 000000000..bb0e0a36d --- /dev/null +++ b/1092/CH9/EX9.5/Example9_5.sce @@ -0,0 +1,41 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 8 ; // Number of poles in the SCIM
+f = 60 ; // Frequency in Hz
+R_r = 0.3 ; // Rotor resistance per phase in ohm
+S_r = 650 ; // Speed in rpm at which motor stalls
+
+// Calculations
+// case a
+S = (120*f)/P ; // Speed in rpm of the rotating magnetic field
+s_b = (S - S_r)/S ; // Breakdown Slip
+
+// case b
+X_lr = R_r / s_b ; // Locked rotor reactance in ohm
+
+// case c
+f_r = s_b * f ; // Rotor frequency in Hz, at the maximum torque point
+
+// case d
+s = 5*(1/100);// Rated slip
+S_r = S * (1 - s); // Full-load in rpm speed at rated slip
+
+// Display the results
+disp("Example 9-5 Solution : ");
+printf(" \n a: S = %d rpm \n s_b = %.3f \n", S , s_b );
+
+printf(" \n b: X_b = %.2f ohm \n ", X_lr );
+
+printf(" \n c: f_r = %.1f Hz \n ", f_r );
+
+printf(" \n d: S = %d rpm \n ", S_r );
+
diff --git a/1092/CH9/EX9.6/Example9_6.sce b/1092/CH9/EX9.6/Example9_6.sce new file mode 100755 index 000000000..90d9834c2 --- /dev/null +++ b/1092/CH9/EX9.6/Example9_6.sce @@ -0,0 +1,67 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 8 ; // Number of poles in the SCIM
+f = 60 ; // Frequency in Hz
+R_r = 0.3 ; // rotor resistance per phase in ohm/phase
+R_x = 0.7 ; // Added resistance in ohm/phase
+R_r_total = R_r + R_x ; // Total resistance per phase in ohm
+S_r = 875 ; // Full-load Speed in rpm
+
+
+// Calculated values from Ex.9-6
+S = 900 ; // Speed in rpm of the rotating magnetic field
+X_lr = 1.08 ; // Locked rotor reactance in ohm
+
+// Calculations
+// case a
+s = (S - S_r)/S ; // Full-load slip,short circuited
+s_r = R_r_total / R_r * s; // New full-load slip with added resistance
+
+S_r_new = S*(1-s_r); // New full-load speed in rpm
+
+// case b
+// Neglecting constant Kn_t ,since we are taking torque ratios
+T_o = ( R_r / ((R_r)^2 + (X_lr)^2) ); // Original torque
+T_f = ( R_r + R_x) / ( (R_r + R_x)^2 + (X_lr)^2 ); // Original torque
+
+torque_ratio = T_f / T_o ; // Ratio of final torque to original torque
+T_final = 2*torque_ratio ;
+
+// Display the results
+disp("Example 9-6 Solution : ");
+printf(" \n a: The full-load slip,short circuited,is ");
+printf(" \n s = %.4f \n",s );
+printf(" \n Since slip is proportional to rotor resistance and since the ");
+printf(" \n increased rotor resistance is R_r = %.1f + %.1f = %d ,",R_x,R_r,R_r_total);
+printf(" \n the new full-load slip with added resistance is : ");
+printf(" \n s_r = %.4f \n",s_r);
+printf(" \n The new full-load speed is : " );
+printf(" \n S(1-s) = %.f rpm \n",S_r_new );
+
+printf(" \n b: The original starting torque T_o was twice the full-load torque");
+printf(" \n with a rotor resistance of %.1f ohm and a rotor reactance of %.2f ohm",R_r,X_lr);
+printf(" \n (Ex.9-5).The new starting torque conditions may be summarized by the ");
+printf(" \n following table and compared from Eq.(9-14),where T_o ");
+printf(" \n is the original torque and T_f is the new torque.");
+
+printf(" \n _________________________________________");
+printf(" \n Condition \t R_r \t X_lr \t T_starting ");
+printf(" \n \t ohm \t ohm \t ");
+printf(" \n _________________________________________");
+printf(" \n Original : \t %.1f \t %.2f \t 2*T_n ",R_r,X_lr);
+printf(" \n New : \t %.1f \t %.2f \t ? ",R_r_total,X_lr);
+printf(" \n _________________________________________\n");
+
+printf(" \n T_o = %.2f * K_n_t",T_o);
+printf(" \n T_f = %.3f * K_n_t",T_f);
+printf(" \n T_f/T_o = %.2f and T_f = %.2f * T_o\n ",torque_ratio,torque_ratio);
+printf(" \n Therefore,\n T_f = %.3f * T_n",T_final);
diff --git a/1092/CH9/EX9.7/Example9_7.sce b/1092/CH9/EX9.7/Example9_7.sce new file mode 100755 index 000000000..af4cfa69f --- /dev/null +++ b/1092/CH9/EX9.7/Example9_7.sce @@ -0,0 +1,54 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 8 ; // Number of poles in the SCIM
+f = 60 ; // Frequency in Hz
+R_r = 0.3 ; // Rotor resistance per phase in ohm
+R_x = 0.7 ; // Added resistance in ohm/phase
+R_r_total = R_r + R_x ; // Total resistance per phase in ohm
+X_lr = 1.08 ; // Locked rotor reactance in ohm
+S_r = 650 ; // Speed in rpm at which motor stalls
+E_lr = 112 ; // Induced voltage per phase
+
+// Calculations
+// case a
+Z_lr = R_r + %i*X_lr ; // Locked rotor impedance per phase
+Z_lr_m = abs(Z_lr);//Z_lr_m = magnitude of Z_lr in ohm
+Z_lr_a = atan(imag(Z_lr) /real(Z_lr))*180/%pi;//Z_lr_a=phase angle of Z_lr in degrees
+
+I_r = E_lr / Z_lr_m ; // Rotor current per phase
+cos_theta_r = cosd(Z_lr_a); // rotor power factor with the rotor short-circuited
+cos_theta = R_r / Z_lr_m ; // rotor power factor with the rotor short-circuited
+
+// case b
+// 1 at the end of Z_lr1 is just used for showing its different form Z_lr
+// and for ease in calculations
+Z_lr1 = R_r_total + %i*X_lr ; // Locked rotor impedance per phase
+Z_lr1_m = abs(Z_lr1);//Z_lr1_m = magnitude of Z_lr1 in ohm
+Z_lr1_a = atan(imag(Z_lr1) /real(Z_lr1))*180/%pi;//Z_lr1_a=phase angle of Z_lr1 in degrees
+
+I_r1 = E_lr / Z_lr1_m ; // Rotor current per phase
+cos_theta_r1 = cosd(Z_lr1_a); // rotor power factor with the rotor short-circuited
+cos_theta1 = R_r_total / Z_lr1_m ; // rotor power factor with the rotor short-circuited
+
+// Display the results
+disp("Example 9-7 Solution : ");
+printf(" \n a: The locked-rotor impedance per phase is : ");
+printf(" \n Z_lr in ohm = "),disp(Z_lr);
+printf(" \n Z_lr = %.2f <%.1f ohm \n",Z_lr_m,Z_lr_a);
+printf(" \n I_r = %.f A \n",I_r);
+printf(" \n cosθ_r = cos(%.1f) = %.3f or \n cosθ = R_r/Z_lr = %.3f",Z_lr_a,cos_theta_r,cos_theta);
+
+printf(" \n\n\n b: The locked-rotor impedance with added rotor resistance per phase is : ");
+printf(" \n Z_lr in ohm = "),disp(Z_lr1);
+printf(" \n Z_lr = %.2f <%.1f ohm \n",Z_lr1_m,Z_lr1_a);
+printf(" \n I_r = %.1f A \n",I_r1);
+printf(" \n cosθ_r = cos(%.1f) = %.3f or \n cosθ = R_r/Z_lr = %.3f",Z_lr1_a,cos_theta_r1,cos_theta1);
diff --git a/1092/CH9/EX9.8/Example9_8.sce b/1092/CH9/EX9.8/Example9_8.sce new file mode 100755 index 000000000..d84dc04d8 --- /dev/null +++ b/1092/CH9/EX9.8/Example9_8.sce @@ -0,0 +1,90 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data (Exs.9-5 through 9-7)
+P = 8 ; // Number of poles in the SCIM
+f = 60 ; // Frequency in Hz
+R_r = 0.3 ; // Rotor resistance per phase in ohm
+X_lr = 1.08 ; // Locked rotor reactance in ohm
+S_r = 650 ; // Speed in rpm at which motor stalls
+E_lr = 112 ; // Induced voltage per phase
+
+disp("Example 9-8 : ");
+printf(" \n The new and the original conditions may be summarized in the following table\n");
+printf(" \n _________________________________________________________");
+printf(" \n Condition \t R_r \t\t X_lr \t\t T_starting ");
+printf(" \n \t ohm \t\t ohm \t ");
+printf(" \n _________________________________________________________");
+printf(" \n Original : \t %.1f \t\t %.2f \t\t T_o = 2*T_n ",R_r,X_lr);
+printf(" \n New :\t(%.1f+R_x) \t %.2f \t\t T_n = 2*T_n ",R_r,X_lr);
+printf(" \n _________________________________________________________\n");
+
+// Calculating
+// case a
+// Neglecting constant Kn_t ,since we are equating torque T_o and T_n
+T_o = ( R_r / ((R_r)^2 + (X_lr)^2) ); // Original torque
+
+// T_o = K_n_t*( 0.3 / ((0.3)^2 + (1.08)^2) );
+// T_n = K_n_t*( 0.3 + R_x) / ( (0.3 + R_x)^2 + (1.08)^2 );
+// T_n = T_o
+// Simplyifing yields
+// 0.3 + R_x = 0.24[(0.3+R_x)^2 + (1.08)^2]
+// Expanding and combining the terms yields
+// 0.24*(R_x)^2 - 0.856*R_x = 0
+// This is a quadratic equation having two roots,which may be factored as
+// R_x*(0.24*R_x - 0.856) = 0,yielding
+// R_x = 0 and R_x = 0.856/0,24 = 3.57
+R_x = poly(0,'R_x'); // Defining a polynomial with variable 'R_x' with root at 0
+a = 0.24 ; // coefficient of x^2
+b = -0.856 ; // coefficient of x
+c = 0 ; // constant
+
+// Roots of p
+R_x1 = ( -b + sqrt (b^2 -4*a*c ) ) /(2* a);
+R_x2=( -b - sqrt (b^2 -4*a*c ) ) /(2* a);
+// Consider R_x>0 value,
+R_x = R_x1;
+
+R_T = R_r + R_x ; // Total rotor resistance in ohm
+
+// case b
+Z_T = R_T + %i*X_lr ; // Total impedance in ohm
+Z_T_m = abs(Z_T);//Z_T_m = magnitude of Z_T in ohm
+Z_T_a = atan(imag(Z_T) /real(Z_T))*180/%pi;//Z_T_a=phase angle of Z_T in degrees
+
+cos_theta = R_T / Z_T_m ; // Rotor PF that will produce the same starting torque
+
+// case c
+Z_r = Z_T_m ; // Impedance in ohm
+I_r = E_lr / Z_r ; // Starting current in A
+
+// Display the results
+disp("Solution : ");
+
+printf(" \n a: T_o = %.2f * K_n_t ",T_o );
+printf(" \n T_n = %.2f * K_n_t \n",T_o );
+printf(" \n Simplyifing yields");
+printf(" \n 0.3 + R_x = 0.24[(0.3+R_x)^2 + (1.08)^2]");
+printf(" \n Expanding and combining the terms yields");
+printf(" \n 0.24*(R_x)^2 - 0.856*R_x = 0");
+printf(" \n This is a quadratic equation having two roots,which may be factored as");
+printf(" \n R_x*(0.24*R_x - 0.856) = 0,yielding");
+printf(" \n R_x = 0 ohm and R_x = 0.856/0.24 = 3.57 ohm\n\n This proves that ");
+printf(" \n Original torque is produced with an external resistance of either ");
+printf(" \n zero or 12 times the origianl rotor resistance.Therefore,\n");
+printf(" \n R_T = R_r + R_x = %.2f ohm \n",R_T);
+
+printf(" \n b: Z_T in ohm = ");disp(Z_T);
+printf(" \n Z_T = %.2f <%.1f ohm ",Z_T_m,Z_T_a);
+printf(" \n cosӨ = R_T / Z_T = %.3f or \n cosӨ = cosd(%.1f) = %.3f\n",cos_theta,Z_T_a,cosd(Z_T_a));
+
+printf(" \n c: I_r = E_lr / Z_r = %.f A \n\n This proves that,",I_r);
+printf(" \n Rotor current at starting is now only 28 percent of the original");
+printf(" \n starting current in part(a) of Ex.9-7");
diff --git a/1092/CH9/EX9.9/Example9_9.sce b/1092/CH9/EX9.9/Example9_9.sce new file mode 100755 index 000000000..647a6926d --- /dev/null +++ b/1092/CH9/EX9.9/Example9_9.sce @@ -0,0 +1,77 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P = 8 ; // Number of poles in the SCIM
+f = 60 ; // Frequency in Hz
+S_r = 875 ; // Full-load Speed in rpm with rotor short-circuited
+R_r = 0.3 ; // rotor resistance per phase in ohm/phase
+R_x = 0.7 ; // Added resistance in ohm/phase
+R_x_a = 1.7 ; // Added resistance in ohm/phase (case a)
+R_x_b = 2.7 ; // Added resistance in ohm/phase (case b)
+R_x_c = 3.7 ; // Added resistance in ohm/phase (case c)
+R_x_d = 4.7 ; // Added resistance in ohm/phase (case d)
+
+// Calculations
+S = (120*f)/P ; // Speed in rpm of the rotating magnetic field
+s_o = (S - S_r)/S ; // Slip at rotor speed 875 rpm
+
+// case a
+s_r_a = s_o * (R_r + R_x_a)/R_r; // Rated slip
+S_r_a = S * (1 - s_r_a); // Full-load speed in rpm for added resistance R_x_a
+
+// case b
+s_r_b = s_o * (R_r + R_x_b)/R_r; // Rated slip
+S_r_b = S * (1 - s_r_b); // Full-load speed in rpm for added resistance R_x_b
+
+// case c
+s_r_c = s_o * (R_r + R_x_c)/R_r; // Rated slip
+S_r_c = S * (1 - s_r_c); // Full-load speed in rpm for added resistance R_x_c
+
+// case d
+s_r_d = s_o * (R_r + R_x_d)/R_r; // Rated slip
+S_r_d = S * (1 - s_r_d); // Full-load speed in rpm for added resistance R_x_d
+
+// Display the results
+disp("Example 9-9 Solution : ");
+
+printf(" \n Slip s_r = s_o*(R_r+R_x)/R_r \n Rotor speed S_r = S_o*(1-s)\n");
+
+printf(" \n Calculated value of s_o = %f , instead of 0.0278(textbook)",s_o)
+printf(" \n so slight variations in the answers below.\n");
+
+printf(" \n a: When R_x = %.1f ohm ",R_x_a);
+printf(" \n s_r = %.3f \n S_r = %.1f rpm \n",s_r_a,S_r_a );
+
+printf(" \n b: When R_x = %.1f ohm ",R_x_b);
+printf(" \n s_r = %.3f \n S_r = %.1f rpm \n",s_r_b,S_r_b );
+
+printf(" \n c: When R_x = %.1f ohm ",R_x_c);
+printf(" \n s_r = %.3f \n S_r = %.1f rpm \n",s_r_c,S_r_c );
+
+printf(" \n d: When R_x = %.1f ohm ",R_x_d);
+printf(" \n s_r = %.3f \n S_r = %.1f rpm \n",s_r_d,S_r_d );
+
+printf(" \n This example,verifies that slip is proportional to rotor resistance");
+printf(" \n as summarized below.");
+
+printf(" \n ___________________________________________________________________");
+printf(" \n R_T(ohm) = R_r+R_x \t\t Slip \t\t Full-load Speed(rpm)");
+printf(" \n ___________________________________________________________________");
+printf(" \n Given \t\t\t Given \t\t Given \t\ ");
+printf(" \n 0.3 \t\t\t 0.0278 \t 875 ");
+printf(" \n 0.3+0.1 = 1.0 \t\t 0.0926 \t 817");
+printf(" \n ___________________________________________________________________");
+printf(" \n Given \t\t\t Calculated \t Calculated \t\ ");
+printf(" \n a. %.1f + %.1f = %.1f \t\t %.3f \t\t %.1f ",R_r,R_x_a,R_r+R_x_a,s_r_a,S_r_a);
+printf(" \n b. %.1f + %.1f = %.1f \t\t %.3f \t\t %.1f ",R_r,R_x_b,R_r+R_x_b,s_r_b,S_r_b);
+printf(" \n c. %.1f + %.1f = %.1f \t\t %.3f \t\t %.1f ",R_r,R_x_c,R_r+R_x_c,s_r_c,S_r_c);
+printf(" \n d. %.1f + %.1f = %.1f \t\t %.3f \t\t %.1f ",R_r,R_x_d,R_r+R_x_d,s_r_d,S_r_d);
+printf(" \n ___________________________________________________________________");
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