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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 1: ELECTROMECHANICAL FUNDAMENTALS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.10: calculate_Z_and_Eg.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"no_of_coils = 40;\n",
+"N = 20; // no of turns in each coil\n",
+"omega = 200; // angular velocity of armature in rad/s\n",
+"phi = 5 * 10 ^ -3; // flux per pole\n",
+"a = 4; // No. of parallel paths\n",
+"P = 4; // No. of poles\n",
+"\n",
+"// Calculations\n",
+"Z = no_of_coils * 2 * N; // No. of conductors\n",
+"\n",
+"E_g = ( phi * Z *  omega * P ) / ( 2 * %pi * a ); // Voltage generated by the \n",
+"// armature between brushes\n",
+"\n",
+"// Display the results\n",
+"disp('Example 1-10 Solution : ');\n",
+"printf('\n Z = % d conductors ', Z);\n",
+"printf('\n Eg = % .2f V between the brushes ', E_g);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.11: calculate_F_and_find_its_direction.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"l = 0.5; // length of the conductor\n",
+"A = 0.1 * 0.2; // area of the pole face\n",
+"phi = 0.5 * 10 ^ -3; // magnetic flux in weber\n",
+"I = 10; //Current in the conductor\n",
+"\n",
+"// Calculations\n",
+"B = ( phi ) / ( A ); // Flux density \n",
+"\n",
+"F = B * I * l; // Magnitude of force\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-11 Solution : ');\n",
+"\n",
+"printf('\n a : F = % .3f N', F );\n",
+"\n",
+"printf('\n b : The force on the conductor is % .3f N in an upward direction as shown in fig 1-13c ', F );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.12: repeat_previous_eg_with_angle_75.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-12\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"l = 0.5; // length of the conductor\n",
+"A = 0.1 * 0.2; // area of the pole face\n",
+"phi = 0.5 * 10 ^ -3; // magnetic flux in weber\n",
+"I = 10; //Current in the conductor\n",
+"theta = 75; // angle between the conductor and the flux density B\n",
+"\n",
+"// Calculations\n",
+"B = ( phi ) / ( A ); // Flux density \n",
+"\n",
+"F = B * I * l * sind(theta); // Magnitude of force\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-12 Solution : ');\n",
+"\n",
+"printf('\n F =% f N in a vertically upward direction ', F );\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.13: calculate_counter_emf.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-13\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"R_a = 0.25; // Armature resistance\n",
+"V_a = 125; // dc bus voltage\n",
+"I_a = 60; // Armature current\n",
+"\n",
+"// Calculations\n",
+"E_c = V_a - I_a * R_a; // Counter EMF generated in the armature conductors of motor\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-13 Solution : ');\n",
+"printf('\n Ec = % d V ', E_c );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.14: calculate_Eg_phi_in_linesperpole_and_mWb.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-13\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_a = 110; // voltage across armature\n",
+"I_a = 60; // Armature current\n",
+"R_a = 0.25; // Armature resistance\n",
+"P = 6; // No. of poles\n",
+"a = 12; // No. of paths\n",
+"Z = 720; // No. of armature conductors\n",
+"S = 1800; // Speed in rpm\n",
+"\n",
+"// Calculations\n",
+"E_g = V_a + I_a * R_a; // Generated EMF in the armature\n",
+" \n",
+"phi_lines = ( E_g * ( 60 * a ) ) / ( ( Z * S * P ) * 10 ^ -8 );\n",
+"// Flux per pole in lines\n",
+"\n",
+"phi_Wb = phi_lines * 10 ^ -8; // Flux per pole in webers\n",
+"\n",
+"// Display the results\n",
+"disp('Example 1-14 Solution : ');\n",
+"\n",
+"printf('\n a : Eg = %d V ', E_g );\n",
+"\n",
+"printf('\n b : phi = %f lines/pole ', phi_lines );\n",
+"\n",
+"printf('\n c : phi = %f Wb ', phi_Wb );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.1: calculate_average_voltage.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"t = 50e-3; // t = time in milli second\n",
+"phi = 8 * 10 ^ 6; // phi = uniform magnetic field in maxwells\n",
+"\n",
+"// Calculations\n",
+"E_av = (phi / t) * 10 ^ -8; // E_av = average voltage generated in the conductor\n",
+"// in volt \n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-1 Solution : ');\n",
+"disp('Average voltage generated in the conductor is : ');\n",
+"printf(' E_av = %.2f V', E_av);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.2: calculate_e_and_E.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"l = 18; // l = length of the conductor in inches\n",
+"B = 50000; // B = uniform magnetic field in lines/sq-inches\n",
+"d = 720; // d = distance travelled by conductor in inches\n",
+"t = 1; // t =time taken for the conductor to  move in second\n",
+"\n",
+"// Calculations\n",
+"v = d/t; // v = velocity in inches/second with which the conductor moves\n",
+"\n",
+"// part a\n",
+"e = B * l * v * 10 ^ -8; // e = instantaneous induced EMF in volt\n",
+"// part b\n",
+"A = d * l; // Area swept by the conductor while moving\n",
+"phi = B * A; // phi = uniform magnetic field \n",
+"E = ( phi / t ) * 10 ^ -8; // E = average induced EMF\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-2 Solution : ');\n",
+"\n",
+"printf(' \n a : e = %.2f V ', e);\n",
+"printf(' \n b : E = %.2f V ', E);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.3: calculate_E.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"l = 18; // l = length of the conductor in inches\n",
+"B = 50000; // B = uniform magnetic field in lines/sq-inches\n",
+"d = 720; // d = distance travelled by conductor in inches\n",
+"t = 1; // t =time taken for the conductor to  move in second\n",
+"theta = 75 // theta = angle between the motion of the conductor and field\n",
+"// in radians\n",
+"\n",
+"// Calculations\n",
+"v = d/t; // v = velocity in inches/second with which the conductor moves\n",
+"\n",
+"E = B * l * v * 10 ^ -8 * sind(theta); // E = Average induced EMF in volt\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-3 Solution : ');\n",
+"\n",
+"disp(' Average induced EMF in volt is :')\n",
+"printf('  E = %.2f V ', E);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.4: calculate_E_for_different_theta.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"v = 1.5; // v = velocity in m/s with which the conductor is moving \n",
+"l = 0.4; // l = length of the conductor\n",
+"B = 1; // B = uniform field intensity in tesla\n",
+"theta_a = 90; // theta_a = angle between the motion of the conductor and field\n",
+"theta_b = 35; // theta_b = angle between the motion of the conductor and field\n",
+"theta_c = 120; // theta_c = angle between the motion of the conductor and field\n",
+"\n",
+"// Calculations\n",
+"E_a = B * l * v * sind(theta_a); // Voltage induced in the conductor for theta_a\n",
+"E_b = B * l * v * sind(theta_b); // Voltage induced in the conductor for theta_b\n",
+"E_c = B * l * v * sind(theta_c); // Voltage induced in the conductor for theta_c\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-1 Solution : ');\n",
+"\n",
+"printf('\n a: E = %.2f V ', E_a);\n",
+"printf('\n b: E = %.3f V ', E_b);\n",
+"printf('\n c: E = %.2f V ', E_c);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.5: calculate_Eperpath_Eg_Ia_Ra_Vt_P.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"no_of_conductors = 40;\n",
+"A = 2; // A = Parallel paths\n",
+"path = A;\n",
+"flux_per_pole = 6.48 * 10 ^ 8; // flux lines \n",
+"S = 30; // S = Speed of the prime mover in rpm\n",
+"R_per_path = 0.01; // Resistance per path\n",
+"I = 10; // Current carried by each condutcor\n",
+"P = 2; // No. of poles\n",
+"\n",
+"// Calculations\n",
+"total_flux = P * flux_per_pole; // Total flux linked in one revolution\n",
+"t = ( 1 / 30 ) * ( 60 ); // time for one revolution\n",
+"\n",
+"e_av_per_conductor = ( total_flux / t ) * 10^-8; // Average voltage generated \n",
+"// per conductor\n",
+"E_path = ( e_av_per_conductor ) * ( no_of_conductors / path ); // Average \n",
+"//voltage  generated per path\n",
+"\n",
+"E_g = E_path; // Generated armature voltage\n",
+"\n",
+"I_a =( I / path ) * ( 2 * path ); // Armature current delivered to an external\n",
+"// load\n",
+"\n",
+"R_a = ( R_per_path) / path * 20; // Armature resistance\n",
+"\n",
+"V_t = E_g - I_a * R_a; // Terminal voltage of generator\n",
+"\n",
+"P = V_t * I_a; // Genrator power rating\n",
+"\n",
+"// Display the results\n",
+"disp('Example 1-5 Solution');\n",
+"\n",
+"printf(' \n a : E/path = %.2f V/path ', E_path );\n",
+"printf(' \n b : Eg = %.2f V ', E_g );\n",
+"printf(' \n c : Ia = %.2f A ', I_a );\n",
+"printf(' \n d : Ra = %.2f ohm ', R_a );\n",
+"printf(' \n e : Vt = %.2f V ', V_t );\n",
+"printf(' \n f : P = %.2f W ', P );\n",
+"\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.6: repeated_previous_eg_with_4poles.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"no_of_conductors = 40;\n",
+"I = 10; // Current carried by each condutcor\n",
+"R_per_path = 0.01; // Resistance per path\n",
+"flux_per_pole = 6.48 * 10 ^ 8; // flux lines \n",
+"P = 2; // No. of poles\n",
+"path = 4; // No. of parallel paths\n",
+"total_flux = P * flux_per_pole; // Total flux linked in one revolution\n",
+"t = 2; // time for one revolution\n",
+"e_av_per_conductor = 6.48; // Average voltage generated per conductor\n",
+"\n",
+"// Calculations\n",
+"E_path = ( e_av_per_conductor ) * ( no_of_conductors / path ); // Average \n",
+"//voltage  generated per path\n",
+" \n",
+"E_g = E_path; // Generated armature voltage\n",
+"\n",
+"I_a =( I / path ) * ( 4 * path ); // Armature current delivered to an external\n",
+"// load \n",
+"\n",
+"R_a = ( ( R_per_path) / path ) * 10; // Armature resistance\n",
+"\n",
+"V_t = E_g - I_a * R_a; // Terminal voltage of generator\n",
+"\n",
+"P = V_t * I_a; // Genrator power rating\n",
+"\n",
+"// Display the results\n",
+"disp('Example 1-6 Solution');\n",
+"\n",
+"printf(' \n a : E/path = %.2f V/path ', E_path );\n",
+"printf(' \n b : Eg = %.2f V ', E_g );\n",
+"printf(' \n c : Ia = %.2f A ', I_a );\n",
+"printf(' \n d : Ra = %.3f ohm ', R_a );\n",
+"printf(' \n e : Vt = %.2f V ', V_t );\n",
+"printf(' \n f : P = %.2f W ', P );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.7: calculate_Eav_per_coil_and_per_coilside.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"N = 1; // no. of turns\n",
+"phi = 6.48 * 10 ^ 8; // Magnetic flux in lines\n",
+"s = 30 / 60; // No. of revolution of the coil per second( refer section 1-14)\n",
+"\n",
+"// Calculations\n",
+"E_av_per_coil = 4 * phi * N * s * 10 ^ -8; // average voltage per coil\n",
+"// for above equation refer section 1-14\n",
+"\n",
+"E_av_per_coil_side = E_av_per_coil * ( 1 / 2); // average voltage per conductor\n",
+"\n",
+"// Display the results\n",
+"disp('Example 1-7 Solution : ')\n",
+"printf(' \n Eav/coil = % .2f V/coil ', E_av_per_coil);\n",
+"printf(' \n Eav/coil side = % .2f V/conductor ', E_av_per_coil_side);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.8: verify_previous_eg_with_phi_in_webers.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"phi_lines = 6.48 * 10 ^ 8; // magnetic flux in lines \n",
+"N = 1; // no. of turns\n",
+"\n",
+"// Calculations\n",
+"phi = phi_lines * 10 ^ -8; // Magnetic flux in weber\n",
+"\n",
+"omega = ( 30 ) * ( 2 * %pi ) * ( 1 / 60 ); // angular velocity in rad/s\n",
+"\n",
+"E_av_per_coil = 0.63662 * omega * phi * N; // average voltage per coil\n",
+"// for the above formula refer section 1-14 eqn (1-4b)\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-8 Solution : ');\n",
+"printf('\n Eav/coil  = % 0.2f V/coil ', E_av_per_coil);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 1.9: verify_eg1_5b_with_eq1_5a.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 1: Electromechanical Fundamentals\n",
+"// Example 1-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 2; // No. of poles\n",
+"Z = 40; // no of conductors\n",
+"a = 2; // a = Parallel paths\n",
+"phi = 6.48 * 10 ^ 8; // magnetic flux\n",
+"S = 30; // Speed of the prime mover\n",
+"\n",
+"// Calculations \n",
+"E_g = ( ( phi * Z * S * P ) / ( 60 * a) ) * 10 ^ -8; // average voltage between \n",
+"// the brushes\n",
+"\n",
+"// Display the result\n",
+"disp('Example 1-9 Solution : ');\n",
+"printf('\n Eg = %.2f V between the brushes ', E_g);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/10-SINGLE_PHASE_MOTORS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/10-SINGLE_PHASE_MOTORS.ipynb
new file mode 100644
index 0000000..d45c5a0
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/10-SINGLE_PHASE_MOTORS.ipynb
@@ -0,0 +1,329 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 10: SINGLE PHASE MOTORS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.1: EX10_1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 10: SINGLE-PHASE MOTORS\n",
+"// Example 10-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"hp = 0.25 ; // Power rating of the single-phase motor in hp\n",
+"V = 110 ; // Voltage rating of the single-phase motor in V\n",
+"I_sw = 4 ; // Starting winding current\n",
+"phi_I_sw = 15 ; // Phase angle in degrees by which I_sw lags behind V\n",
+"I_rw = 6 ; // Running winding current\n",
+"phi_I_rw = 40 ; // Phase angle in degrees by which I_rw lags behind V \n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"I_s = I_sw * exp( %i * -phi_I_sw*(%pi/180) ); //  starting current in A\n",
+"// (%pi/180) for degrees to radians conversion of phase angle\n",
+"I_s_m = abs(I_s);//I_s_m = magnitude of I_s in A\n",
+"I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees\n",
+"\n",
+"I_r = I_rw * exp( %i * -phi_I_rw*(%pi/180) ); //  running current in A\n",
+"I_r_m = abs(I_r);//I_r_m = magnitude of I_r in A\n",
+"I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees\n",
+"\n",
+"I_t = I_s + I_r ; // Total starting current in A\n",
+"I_t_m = abs(I_t);//I_t_m = magnitude of I_t in A\n",
+"I_t_a = atan(imag(I_t) /real(I_t))*180/%pi;//I_t_a=phase angle of I_t in degrees\n",
+"Power_factor = cosd(I_t_a); // Power factor\n",
+"\n",
+"// case b\n",
+"Is_cos_theta = real(I_s); // Component of the starting winding current in phase \n",
+"// with the supply voltage in  A\n",
+"\n",
+"// case c\n",
+"Ir_sin_theta = imag(I_r); // Component of the running winding current that lags\n",
+"// the supply voltage by 90 degrees\n",
+"\n",
+"// case d\n",
+"phase = ( phi_I_rw - phi_I_sw ); // Phase angle between the starting and running \n",
+"// currents in degrees\n",
+"\n",
+"// Display the results\n",
+"disp('Example 10-1 Solution : ');\n",
+"printf(' \n a: I_s = %d <-%d A ', I_sw , phi_I_sw );\n",
+"printf(' \n    I_s in A =  ' );disp(I_s);\n",
+"printf(' \n    I_r = %d <-%d A ', I_rw , phi_I_rw );\n",
+"printf(' \n    I_r in A =  ' );disp(I_r);\n",
+"printf(' \n    I_t in A =  ' );disp(I_t);\n",
+"printf(' \n    I_t = %.2f <%d A ', I_t_m , I_t_a );\n",
+"printf(' \n\n    Power factor = cos(%d) = %.3f lagging \n', I_t_a ,Power_factor);\n",
+"\n",
+"printf(' \n b: Is*cosθ = %.2f A  (from a)\n ', Is_cos_theta );\n",
+"\n",
+"printf(' \n c: (from a),\n    Ir*sinθ in A = ' );disp(%i*Ir_sin_theta);\n",
+"\n",
+"printf(' \n d: (θ_r - θ_s) = %d degrees ', phase);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.2: calculate_Ps_Pr_Pt_and_motor_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 10: SINGLE-PHASE MOTORS\n",
+"// Example 10-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data as per Ex.10-1\n",
+"hp = 0.25 ; // Power rating of the single-phase motor in hp\n",
+"V = 110 ; // Voltage rating of the single-phase motor in V\n",
+"I_s = 4 ; // Starting winding current\n",
+"phi_I_s = 15 ; // Phase angle in degrees by which I_s lags behind V\n",
+"I_r = 6 ; // Running winding current\n",
+"phi_I_r = 40 ; // Phase angle in degrees by which I_r lags behind V \n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"P_s = V * I_s * cosd(phi_I_s); // Power dissipated in the starting winding in W\n",
+"\n",
+"// case b\n",
+"P_r = V * I_r * cosd(phi_I_r); // Power dissipated in the running winding in W\n",
+"\n",
+"// case c\n",
+"P_t = P_s + P_r ; // Total instantaneous power dissipated during starting in W\n",
+"\n",
+"// case d\n",
+"P_r_d = P_r ; // Total steady-state power dissipated during running in W\n",
+"\n",
+"// case e\n",
+"eta = ( hp * 746 ) / P_r * 100 ; // Motor efficiency in percent\n",
+"\n",
+"// Display the results\n",
+"disp('Example 10-2 Solution : ');\n",
+"printf(' \n a: Power dissipated in the starting winding\n    P_s = %d W \n', P_s );\n",
+"\n",
+"printf(' \n b: Power dissipated in the running winding\n    P_r = %.1f W \n', P_r );\n",
+"\n",
+"printf(' \n c: Total instantaneous power dissipated during starting\n    P_t = %.1f W \n', P_t );\n",
+"\n",
+"printf(' \n d: Total steady-state power dissipated during running\n    P_r = %.1f W \n', P_r_d );\n",
+"\n",
+"printf(' \n e: Motor efficiency \n    η = %.f percent \n', eta );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.3: calculate_total_starting_current_and_sine_of_angle_between_Is_Ir.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 10: SINGLE-PHASE MOTORS\n",
+"// Example 10-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"hp = 0.25 ; // Power rating of the single-phase motor in hp\n",
+"V = 110 ; // Voltage rating of the single-phase motor in V\n",
+"I_sw = 4 ; // Starting winding current\n",
+"phi_I_sw = 15 ; // Phase angle in degrees by which I_sw lags behind V\n",
+"I_rw = 6 ; // Running winding current\n",
+"phi_I_rw = 40 ; // Phase angle in degrees by which I_rw lags behind V \n",
+"// when the capacitor is added to the auxiliary starting winding of the motor\n",
+"// of Ex.10-1 , I_s leads V by 42 degrees so,\n",
+"phi_I_sw_new = 42 ; // I_s leads V by phi_I_sw_new degrees\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"I_s = I_sw * exp( %i * phi_I_sw_new*(%pi/180) ); //  starting current in A\n",
+"// (%pi/180) for degrees to radians conversion of phase angle\n",
+"I_s_m = abs(I_s);//I_s_m = magnitude of I_s in A\n",
+"I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees\n",
+"\n",
+"I_r = I_rw * exp( %i * -phi_I_rw*(%pi/180) ); //  running current in A\n",
+"I_r_m = abs(I_r);//I_r_m = magnitude of I_r in A\n",
+"I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees\n",
+"\n",
+"I_t = I_s + I_r ; // Total starting current in A\n",
+"I_t_m = abs(I_t);//I_t_m = magnitude of I_t in A\n",
+"I_t_a = atan(imag(I_t) /real(I_t))*180/%pi;//I_t_a=phase angle of I_t in degrees\n",
+"Power_factor = cosd(I_t_a); // Power factor\n",
+"\n",
+"// case b\n",
+"theta = ( phi_I_rw - (-phi_I_sw_new) ); \n",
+"sin_theta = sind(theta);// Sine of the angle between the \n",
+"// starting and running currents \n",
+"phase = 25 ; // Phase angle between the starting and running \n",
+"// currents in degrees (from Ex.10-1)\n",
+"\n",
+"// case c\n",
+"// Ratio of starting torques (capacitor to resistance start)\n",
+"ratio_T = sind(theta) / sind(phase); \n",
+"\n",
+"// Display the results\n",
+"disp('Example 10-3 Solution : ');\n",
+"printf(' \n a: I_s = %d <%d A ', I_sw , phi_I_sw_new );\n",
+"printf(' \n    I_s in A =  ' );disp(I_s);\n",
+"printf(' \n    I_r = %d <-%d A ', I_rw , phi_I_rw );\n",
+"printf(' \n    I_r in A =  ' );disp(I_r);\n",
+"printf(' \n    I_t in A =  ' );disp(I_t);\n",
+"printf(' \n    I_t = %.2f <%.1f A ', I_t_m , I_t_a );\n",
+"printf(' \n\n    Power factor = cos(%.1f) = %.3f lagging \n', I_t_a ,Power_factor);\n",
+"\n",
+"printf(' \n b: sin(%d - (-%d)) = sin(%d) = %.4f\n',phi_I_rw,phi_I_sw_new,theta,sin_theta);\n",
+"\n",
+"printf(' \n c: The steady state starting current has been reduced from');\n",
+"printf(' \n    9.77 <-30 A to %.2f <%.1f A ,',I_t_m ,I_t_a );\n",
+"printf(' \n    and the power factor has risen from 0.866 lagging to %.3f.',Power_factor);\n",
+"printf(' \n    The motor develops maximum starting torque(T = K*I_b*ϕ*cosθ) with');\n",
+"printf(' \n    minimum starting current.The ratio of starting torques ');\n",
+"printf(' \n    (capacitor to resistance start) is : \n');\n",
+"printf(' \n    T_cs/T_rs = sin(%d)/sin(%d) = %.3f',theta,phase,ratio_T)"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.4: calculate_ratios_of_T_and_efficiency_and_rated_PF_and_hp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 10: SINGLE-PHASE MOTORS\n",
+"// Example 10-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data (from Table 10-2)\n",
+"T_r = 1 ; // Rated torque in lb-ft\n",
+"T_s = 4.5 ; // Starting torque in lb-ft (rfom Locked-Rotor Data)\n",
+"T_br = 2.5 ; // Breakdown torque in lb-ft (Breakdown-Torque Data)\n",
+"\n",
+"// Rated Load Data\n",
+"P = 400 ; // Rated input power in W\n",
+"V = 115 ; // Rated input voltage in volt\n",
+"I_t = 5.35 ; // Rated input current in A\n",
+"Speed = 1750 ; // Rated speed in rpm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"ratio_s_r_T = T_s / T_r ; // Ratio of starting to rated torque\n",
+"\n",
+"// case b\n",
+"ratio_s_br_T = T_br / T_r ; // Ratio of breakdown to rated torque\n",
+"\n",
+"// case c\n",
+"P_o_hp = 1 / 3 ; // Power output in hp\n",
+"P_o = P_o_hp * 746 ; // Power output in W \n",
+"eta = P_o / P * 100 ; // Rated load efficiency\n",
+"\n",
+"// case d\n",
+"S = V * I_t ; // VA rating of the motor\n",
+"cos_theta = P / S ; // Rated load - power factor\n",
+"\n",
+"// case e\n",
+"T = 1 ; // Rated load torque in lb-ft\n",
+"hp = (T*Speed)/5252 ; // Rated load horsepower\n",
+"\n",
+"// Display the results\n",
+"disp('Example 10-4 Solution : ');\n",
+"\n",
+"printf(' \n a: T_s/T_r = %.1f \n ',ratio_s_r_T );\n",
+"\n",
+"printf(' \n b: T_br/T_r = %.1f \n ',ratio_s_br_T );\n",
+"\n",
+"printf(' \n c: Rated load efficiency \n    η = %.1f percent \n ',eta );\n",
+"\n",
+"printf(' \n d: Rated load power factor\n    cosθ = %.4f \n ',cos_theta );\n",
+"\n",
+"printf(' \n e: Rated load horsepower\n    hp = %.4f hp ',hp);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/11-SPECIALIZED_DYNAMOS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/11-SPECIALIZED_DYNAMOS.ipynb
new file mode 100644
index 0000000..03ef31f
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/11-SPECIALIZED_DYNAMOS.ipynb
@@ -0,0 +1,279 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 11: SPECIALIZED DYNAMOS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.1: calculate_S_V_P_T_A_and_B_from_torque_speed_relations_fig.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 11: SPECIALIZED DYNAMOS\n",
+"// Example 11-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Torque - speed relations shown in Fig.11-3b for a dc servomotor.\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// Extrapolating to load line point x,\n",
+"S = 800 ; // Motor speed at point x\n",
+"V = 60 ; // Armature voltage in volt at point x\n",
+"\n",
+"// case b\n",
+"// At standstill, 60 V yields 4.5 lb-ft of starting torque\n",
+"T = 4.5 ;\n",
+"\n",
+"// case c\n",
+"P_c = (T*S)/5252 ; // Power delivered to the load in hp (from case a conditions)\n",
+"P_c_watt = P_c * 746 ; // P_c in W\n",
+"// case d\n",
+"// At point o:\n",
+"T_d = 1.1 ; // Starting torque in lb-ft (subscript d indicates case d) and\n",
+"S_d = 410 ; // Motor speed at point at point o\n",
+"\n",
+"// case e\n",
+"// At point w:\n",
+"T_e = 2.4 ; // Starting torque in lb-ft (subscript e indicates case e) and\n",
+"S_e = 900 ; // Motor speed at point at point w\n",
+"\n",
+"// case f\n",
+"P_d = (T_d*S_d)/5252 ; // Power delivered to the load in hp (from case d conditions)\n",
+"P_d_watt = P_d * 746 ; // P_d in W\n",
+"\n",
+"// case g\n",
+"P_f = (T_e*S_e)/5252 ; // Power delivered to the load in hp (from case f conditions)\n",
+"P_f_watt = P_f * 746 ; // P_f in W\n",
+"\n",
+"// case h\n",
+"// Upper limit of power ranges A and B are:\n",
+"A = 65 ; // Upper limit of power range A in W\n",
+"B = 305 ; // Upper limit of power range B in W\n",
+"\n",
+"// Display the results\n",
+"disp('Example 11-1 Solution : ');\n",
+"\n",
+"printf(' \n a: Extrapolating to load line point x,\n    S = %d rpm ',S);\n",
+"printf(' \n    Load line voltage is %d V \n',V);\n",
+"\n",
+"printf(' \n b: At standstill, %d V yields T = %.1f lb-ft of starting torque\n',V,T);\n",
+"\n",
+"printf(' \n c: Power delivered to the load in hp (from case a conditions)');\n",
+"printf(' \n    P = %.4f hp = %d W \n',P_c,P_c_watt);\n",
+"\n",
+"printf(' \n d: At point o:\n    T = %.1f lb-ft and S = %d rpm \n',T_d,S_d);\n",
+"\n",
+"printf(' \n e: At point w:\n    T = %.1f lb-ft and S = %d rpm \n',T_e,S_e);\n",
+"\n",
+"printf(' \n f: P = %.4f hp = %.1f W \n ',P_d,P_d_watt);\n",
+"\n",
+"printf(' \n g: P = %.4f hp = %.f W \n',P_f,P_f_watt );\n",
+"\n",
+"printf(' \n h: A = %d W  and  B = %d W ', A , B );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.2: calculate_stepping_angle.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 11: SPECIALIZED DYNAMOS\n",
+"// Example 11-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// VR stepper motor\n",
+"n = 3 ; // Number of stacks or phases\n",
+"P_a = 16 ; // Number of rotor teeth (subscript a indicates case a)\n",
+"// PM stepper\n",
+"P_b = 24 ; // Number of poles (subscript b indicates case b)\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"alpha_a = 360 / (n*P_a); // Stepping angle in degrees per step \n",
+"\n",
+"alpha_b = 360 / (n*P_b); // Stepping angle in degrees per step \n",
+"\n",
+"// Display the results\n",
+"disp('Example 11-2 Solution : ');\n",
+"printf(' \n a: alpha α = %.1f degrees/step \n ', alpha_a );\n",
+"\n",
+"printf(' \n b: alpha α = %.1f degrees/step \n ', alpha_b );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.3: calculate_stepping_length.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 11: SPECIALIZED DYNAMOS\n",
+"// Example 11-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Hybrid stepping motor\n",
+"P = 50 ; // Number of rotor teeth\n",
+"\n",
+"// Calculation\n",
+"\n",
+"alpha = 90 / P ; // Stepping angle in degrees\n",
+"\n",
+"// Display the result\n",
+"disp('Example 11-3 Solution : ');\n",
+"printf(' \n alpha α = %.1f degrees ', alpha );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.4: calculate_synchronous_velocity.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 11: SPECIALIZED DYNAMOS\n",
+"// Example 11-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"tou = 0.1 ; // Pole pitch of a double-sided primary LIM in meter\n",
+"f = 60 ; // Frequency applied to the primary LIM in Hz\n",
+"\n",
+"// Calculation\n",
+"v_s = 2 * f * tou ; // Synchronous velocity in meter/second\n",
+"\n",
+"// Display the result\n",
+"disp('Example 11-4 Solution : ');\n",
+"printf(' \n Synchronous velocity : \n v_s = %d m/s ', v_s );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 11.5: calculate_slip_of_DSLIM.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 11: SPECIALIZED DYNAMOS\n",
+"// Example 11-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"v_s = 12 ; // Synchronous velocity in meter/second\n",
+"v = 10 ; // Secondary sheet in Ex.11-4 moves at a linear velocity in m/s\n",
+"\n",
+"// Calculation\n",
+"s = (v_s - v)/v_s ; // Slip of the DSLIM\n",
+"\n",
+"// Display the result\n",
+"disp('Example 11-5 Solution : ');disp('From Eq.(11-5)')\n",
+"printf(' \n Slip of the DSLIM : \n s = %.3f  ',s );"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/12-POWER_ENERGY_AND_EFFICIENCY_RELATIONS_OF_DC_AND_AC_DYNAMOS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/12-POWER_ENERGY_AND_EFFICIENCY_RELATIONS_OF_DC_AND_AC_DYNAMOS.ipynb
new file mode 100644
index 0000000..e5cc364
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/12-POWER_ENERGY_AND_EFFICIENCY_RELATIONS_OF_DC_AND_AC_DYNAMOS.ipynb
@@ -0,0 +1,1897 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 12: POWER ENERGY AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.10: efficiency_Pf_Pd_Pr_Ia_LF_max_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 125 ; // Voltage rating of genrator in volt\n",
+"P_o = 12500 ; // Power rating of genrator in W\n",
+"P_hp = 20 ; // Power rating of motor in hp\n",
+"R_a = 0.1 ; // Armture resistance in ohm\n",
+"R_f = 62.5 ; // Field resistance in ohm\n",
+"P_var = 1040 ; // Rated variable electric loss in W\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"P_in = P_hp * 746 ; // Power input to generator in W\n",
+"eta = P_o / P_in * 100 ; // Efficiency\n",
+"\n",
+"// case b\n",
+"V_f = V  ; // Voltage across shunt field wdg in volt\n",
+"P_sh_loss = (V_f)^2 / R_f ; // Shunt field loss in W\n",
+"\n",
+"// case c\n",
+"V_L = V ; \n",
+"I_L = P_o / V_L ; // Line current in A\n",
+"I_f = V_f / R_f ; // Field current in A\n",
+"I_a = I_L + I_f ; // Armature current in A\n",
+"E_g = V_L + I_a*R_a ; // Generated EMF in volt\n",
+"\n",
+"P_d1 = E_g * I_a ; // Generated electric power in W\n",
+"P_f = V_f * I_f ;\n",
+"P_d2 = P_o + P_var + P_f ; // Generated electric power in W\n",
+"\n",
+"// case d\n",
+"P_d = P_d1;\n",
+"P_r = P_in - P_d ; // Rotational power losses in W\n",
+"\n",
+"// case e\n",
+"P_k = P_r + V_f*I_f ; // Constant losses in W\n",
+"Ia = sqrt(P_k/R_a); // Armature current in A for max.efficiency\n",
+"\n",
+"// case f\n",
+"I_a_rated = I_a ; // Rated armature current in A\n",
+"LF = Ia / I_a ; // Load fraction\n",
+"\n",
+"// case g\n",
+"rated_output = 12500 ; // Rated output in kW\n",
+"// Maximum efficiency\n",
+"eta_max = ( LF * rated_output ) / ( ( LF * rated_output ) + (2*P_k) ) * 100 ;\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-10 Solution : ');\n",
+"\n",
+"printf(' \n a: Efficiency :\n    η = %f percent ≃ %.1f percent \n ',eta,eta);\n",
+"\n",
+"printf(' \n b: Shunt field loss :\n    (V_f)^2/R_f = %d W \n ',P_sh_loss);\n",
+"\n",
+"printf(' \n c: Line current : I_L = %d A \n\n    Field current : I_f = %d A',I_L,I_f);\n",
+"printf(' \n\n    Armature current : I_a = %d A ',I_a);\n",
+"printf(' \n\n    Generated EMF : E_g = %.1f V ',E_g);\n",
+"printf(' \n\n    Generated electric power : ');\n",
+"printf(' \n    1. P_d = %d W \n\n    2. P_d = %d W \n ',P_d1,P_d2);\n",
+"\n",
+"printf(' \n d: Rotational power losses :\n    P_r = %f W ≃ %.f W \n',P_r,P_r);\n",
+"\n",
+"printf(' \n e: Constant losses : P_k = %f W ≃ %.f W \n ', P_k ,P_k);\n",
+"printf(' \n    Armature current for max.efficiency : I_a = %.1f A \n ',Ia);\n",
+"\n",
+"printf(' \n f: Load fraction : L.F. = %.2f \n ',LF);\n",
+"\n",
+"printf(' \n g: Maximum efficiency : η = %f percent ≃ %.2f percent',eta_max,eta_max);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.11: efficiency_at_different_LF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data (from Ex.12-10)\n",
+"V = 125 ; // Voltage rating of genrator in volt\n",
+"P_o = 12500 ; // Power rating of genrator in W\n",
+"P_hp = 20 ; // Power rating of motor in hp\n",
+"R_a = 0.1 ; // Armture resistance in ohm\n",
+"R_f = 62.5 ; // Field resistance in ohm\n",
+"P_var = 1040 ; // Rated variable electric loss in W\n",
+"\n",
+"// Calculated data from Ex.12-10\n",
+"P_k = 1380 ; // Constant losses in W\n",
+"\n",
+"// Calculations\n",
+"// Efficiency of the dc shunt generator\n",
+"// η = (output*L.F) / ( (output*L.F) + P_k + (L.F)^2 * P_a_rated ) * 100 \n",
+"output = P_o ;\n",
+"P_a_rated = P_var ;\n",
+"\n",
+"// case a\n",
+"LF1 = 25*(1/100); // At 25 % rated output\n",
+"// Efficiency of the dc shunt generator at 25 % rated output\n",
+"eta_1 = (output*LF1) / ( (output*LF1) + P_k + (LF1)^2 * P_a_rated ) * 100 ; \n",
+"\n",
+"// case b\n",
+"LF2 = 50*(1/100); // At 50 % rated output\n",
+"// Efficiency of the dc shunt generator at 50 % rated output\n",
+"eta_2 = (output*LF2) / ( (output*LF2) + P_k + (LF2)^2 * P_a_rated ) * 100 ; \n",
+"\n",
+"// case c\n",
+"LF3 = 75*(1/100); // At 75 % rated output\n",
+"// Efficiency of the dc shunt generator at 75 % rated output\n",
+"eta_3 = (output*LF3) / ( (output*LF3) + P_k + (LF3)^2 * P_a_rated ) * 100 ; \n",
+"\n",
+"// case d\n",
+"LF4 = 125*(1/100); // At 125 % rated output\n",
+"// Efficiency of the dc shunt generator at 125 % rated output\n",
+"eta_4 = (output*LF4) / ( (output*LF4) + P_k + (LF4)^2 * P_a_rated ) * 100 ; \n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-11 Solution : ');\n",
+"\n",
+"printf(' \n a: η at %.2f rated output = %.2f percent \n ',LF1,eta_1);\n",
+"\n",
+"printf(' \n b: η at %.2f rated output = %.2f percent \n ',LF2,eta_2);\n",
+"printf(' \n    Please note: Calculation error for case b: η in the textbook.\n');\n",
+"\n",
+"printf(' \n c: η at %.2f rated output = %.2f percent \n ',LF3,eta_3);\n",
+"\n",
+"printf(' \n d: η at %.2f rated output = %.2f percent \n ',LF4,eta_4);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.12: Ia_Ra_Pf_Pk_Pcu_efficiencies_Pd.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-12\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3-phase Y-connected alternator\n",
+"kVA = 100 ; // kVA rating of the alternator\n",
+"V  = 1100 ; // Rated voltage of the alternator in volt\n",
+"I_a_nl = 8 ; // No-load armature current in A\n",
+"P_in_nl = 6000 ; // No-load Power input to the armature in W\n",
+"V_oc = 1350 ; // Open-ckt line voltage in volt\n",
+"I_f = 18 ; // Field current in A\n",
+"V_f = 125 ; // voltage across field winding in volt\n",
+"\n",
+"// Calculations\n",
+"// From Ex.6-4,\n",
+"R_a = 0.45 ; // Armature resistance in ohm/phase\n",
+"I_a_rated = 52.5 ; // Rated armature current in A/phase \n",
+"\n",
+"// case a\n",
+"P_r = P_in_nl - 3 * (I_a_nl)^2 * R_a ; // Rotational loss of synchronous dynamo in W\n",
+"\n",
+"// case b\n",
+"P_f = V_f*I_f ; // Field copper loss in W\n",
+"\n",
+"// case c\n",
+"P_k = P_r + P_f ; // Fixed losses in W at rated synchronous speed\n",
+"Pk  = P_k / 1000 ; // Fixed losses in kW at rated synchronous speed\n",
+"\n",
+"// case d\n",
+"P_cu = 3 * (I_a_rated)^2 * R_a ; // Rated electric armature cu-loss in W\n",
+"P_cu_kW = P_cu / 1000 ; // Rated electric armature cu-loss in kW\n",
+"\n",
+"LF1 = 1/4 ; // Load fraction\n",
+"LF2 = 1/2 ; // Load fraction\n",
+"LF3 = 3/4 ; // Load fraction\n",
+"P_cu_LF1 = P_cu * (LF1)^2 ; // Electric armature cu-loss in W at 1/4 load\n",
+"P_cu_LF2 = P_cu * (LF2)^2 ; // Electric armature cu-loss in W at 1/2 load\n",
+"P_cu_LF3 = P_cu * (LF3)^2 ; // Electric armature cu-loss in W at 3/4 load\n",
+"\n",
+"P_cu_LF1_kW = P_cu_LF1 / 1000 ; // Electric armature cu-loss in kW at 1/4 load\n",
+"P_cu_LF2_kW = P_cu_LF2 / 1000 ; // Electric armature cu-loss in kW at 1/2 load\n",
+"P_cu_LF3_kW = P_cu_LF3 / 1000 ; // Electric armature cu-loss in kW at 3/4 load\n",
+"\n",
+"\n",
+"// case e\n",
+"PF = 0.9 ; // Power factor lagging\n",
+"// Efficiency\n",
+"// η = LF(rated kVA)*PF / ( LF(rated kVA)*PF + P_k + P_cu ) * 100 \n",
+"eta_1 = (LF1 * kVA * PF) / ( (LF1 * kVA * PF) + Pk + P_cu_LF1_kW ) * 100 ;// Efficiency at 1/4 load\n",
+"eta_2 = (LF2 * kVA * PF) / ( (LF2 * kVA * PF) + Pk + P_cu_LF2_kW ) * 100 ;// Efficiency at 1/2 load\n",
+"eta_3 = (LF3 * kVA * PF) / ( (LF3 * kVA * PF) + Pk + P_cu_LF3_kW ) * 100 ;// Efficiency at 3/4 load\n",
+"eta_fl = (kVA * PF) / ( (kVA * PF) + Pk + P_cu_kW ) * 100 ;// Efficiency at full load\n",
+"\n",
+"// case f\n",
+"Ia = sqrt(P_k/(3*R_a)); // Armature current in A for max.efficiency at 0.9 PF lagging\n",
+"LF = Ia / I_a_rated ; // Load fraction for max.efficiency\n",
+"// at max.efficiency P_cu = P_k\n",
+"eta_max = (LF * kVA * PF) / ( (LF * kVA * PF) + 2*Pk ) * 100 ;// Max Efficiency 0.9 PF lagging \n",
+"\n",
+"// case g\n",
+"P_o = kVA*PF ; // Output power at 0.9 PF lagging\n",
+"I_a = I_a_rated ;\n",
+"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\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-12 Solution : ');\n",
+"\n",
+"printf(' \n From Ex.6-4,\n R_a = %.2f Ω/phase',R_a);\n",
+"printf(' \n I_a(rated) = %.1f A \n ',I_a_rated);\n",
+"\n",
+"printf(' \n a: Rotational loss of synchronous dynamo :\n    P_r = %.f W \n',P_r);\n",
+"\n",
+"printf(' \n b: Field copper loss :\n    P_f = %d W \n ',P_f);\n",
+"\n",
+"printf(' \n c: Fixed losses at rated synchronous speed :\n    P_k = %.f W\n',P_k);\n",
+"\n",
+"printf(' \n d: P_cu at rated load = %.f W\n    P_cu ,',P_cu);\n",
+"printf(' \n    at %.2f rated load = %.1f W',LF1 , P_cu_LF1);\n",
+"printf(' \n    at %.2f rated load = %.1f W',LF2 , P_cu_LF2);\n",
+"printf(' \n    at %.2f rated load = %.1f W \n',LF3 , P_cu_LF3);\n",
+"\n",
+"\n",
+"printf(' \n e: Efficiency :\n    η at %.2f load = %.1f percent',LF1,eta_1);\n",
+"printf(' \n    η at %.2f load = %.1f percent',LF2,eta_2);\n",
+"printf(' \n    η at %.2f load = %.1f percent',LF3,eta_3);\n",
+"printf(' \n    η at full-load = %.1f percent \n',eta_fl);\n",
+"\n",
+"printf(' \n f: Armature current for max.efficiency at 0.9 PF lagging :');\n",
+"printf(' \n    I_a(max) = %f A ≃ %.1f A\n',Ia,Ia);\n",
+"printf(' \n    L.F. = %.2f \n',LF);\n",
+"printf(' \n    Maximum efficiency :\n    η_max = %.1f percent \n ',eta_max);\n",
+"\n",
+"printf(' \n g: Armature power developed at 0.9 PF lagging at full-load :');\n",
+"printf(' \n    P_d = %.2f kW ',P_d);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.13: Pf_Pcu_Zs_VR_efficiencies_Pd.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-13\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3-phase Y-connected alternator\n",
+"kVA = 1000 ; // kVA rating of the alternator\n",
+"V  = 2300 ; // Rated voltage of the alternator in volt\n",
+"\n",
+"// DC MOTOR\n",
+"P_hp = 100 ; // Power rating of the dc motor in hp\n",
+"V_motor  = 240 ; // Rated voltage of the motor in volt\n",
+"// 4-step efficiency/regulation test\n",
+"// Test 1 \n",
+"P_1 = 7.5 ; // motor output in kW\n",
+"\n",
+"// Test 2\n",
+"P_2 = 16 ; // motor output in kW\n",
+"VfIf = 14 ; // Field losses in kW\n",
+"P_f = VfIf ; // Field losses in kW\n",
+"\n",
+"// Test 3\n",
+"P_3 = 64.2 ; // motor output in kW\n",
+"I_sc = 251 ; // Short ckt current in A\n",
+"\n",
+"// Test 4\n",
+"V_L = 1443 ; // Line voltage in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"P_r = P_2 ; // Rotational losses in kW From test 2\n",
+"\n",
+"// case b\n",
+"P_cu = P_3 - P_1 ; // Full-load armature copper loss in kW\n",
+"\n",
+"// case c\n",
+"E_gL = V_L ; // Generated line voltage in volt\n",
+"Z_s = (E_gL/sqrt(3)) / I_sc ; // Synchronous impedance of the armature in ohm\n",
+"\n",
+"// case d\n",
+"R_a = 0.3 ; // Armature resistance in ohm\n",
+"X_s = sqrt( (Z_s)^2 - (R_a)^2 );  // Synchronous reactance of the armature in ohm\n",
+"\n",
+"// case e\n",
+"cos_theta = 0.8 ; // PF lagging\n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"V_p = V / sqrt(3); // Phase voltage in volt\n",
+"\n",
+"// Generated voltage per phase in volt\n",
+"I_a = I_sc ; // Armature current in A\n",
+"\n",
+"E_gp = (V_p*cos_theta + I_a*R_a) + %i*(V_p*sin_theta + I_a*X_s);\n",
+"E_gp_m = abs(E_gp);//E_gp_m=magnitude of E_gp in volt\n",
+"E_gp_a = atan(imag(E_gp) /real(E_gp))*180/%pi;//E_gp_a=phase angle of E_gp in degrees\n",
+"\n",
+"V_nl = E_gp_m ; // No-load voltage in volt\n",
+"V_fl = V_p ; // Full-load voltage in volt\n",
+"\n",
+"VR = (V_nl - V_fl)/V_fl * 100 ; // Alternator voltage regulation\n",
+"\n",
+"// case f\n",
+"PF = 0.8 ; // lagging PF\n",
+"LF = 1 ; // load fraction\n",
+"eta_rated = (LF*kVA*PF)/( (LF*kVA*PF) + (P_f + P_r) + P_cu ) * 100 ; // Efficiency at 0.8 lagging PF\n",
+"\n",
+"// case g\n",
+"P_k = (P_f + P_r) ; // Constant losses in kW\n",
+"L_F = sqrt(P_k/P_cu); // Load fraction for max.efficiency\n",
+"// at max.efficiency P_k = P_cu\n",
+"eta_max = (L_F*kVA*PF)/( (L_F*kVA*PF) + 2*P_k ) * 100 ; // Max.Efficiency at 0.8 lagging PF\n",
+"\n",
+"\n",
+"// case h\n",
+"P_o = kVA ; // Output power in kVA\n",
+"P_d = P_o +(3*(I_a)^2*R_a/1000) + (VfIf) ; // Armature power developed in kW at unity PF at rated-load\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-13 Solution : ');\n",
+"\n",
+"printf(' \n a: From Test 2, Rotational losses :\n    P_r = %d kW \n',P_r);\n",
+"\n",
+"printf(' \n b: Full-load armature copper loss :\n    P_cu = %.1f kW \n',P_cu);\n",
+"\n",
+"printf(' \n c: Synchronous impedance of the armature :\n    Z_s = %f Ω ≃ %.2f Ω  \n',Z_s,Z_s);\n",
+"\n",
+"printf(' \n d: Synchronous reactance of the armature :\n    jX_s = %f Ω ≃ %.2f Ω \n',X_s,X_s);\n",
+"\n",
+"printf(' \n e: E_gp = ');disp(E_gp);\n",
+"printf(' \n    E_gp = %.f <%.1f V\n',E_gp_m,E_gp_a);\n",
+"printf(' \n    Alternator voltage regulation :\n    VR = %.2f percent \n',VR);\n",
+"\n",
+"printf(' \n    Obtained VR value through scilab calculation is slightly different from textbook');\n",
+"printf(' \n    because of non-approximation of Z_s,X_s and E_gp while calculating in scilab.\n');\n",
+"\n",
+"printf(' \n f: Alternator efficiency at 0.8 lagging PF :\n    η_rated = %.1f percent\n',eta_rated);\n",
+"\n",
+"printf(' \n g: L.F = %.4f\n',L_F);\n",
+"printf(' \n    Max.Efficiency at 0.8 lagging PF :\    η_max = %.2f percent \n',eta_max );\n",
+"\n",
+"printf(' \n h: Power developed by the alternator armature at rated load,unity PF :');\n",
+"printf(' \n    P_d = %.f kW',P_d);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.14: Pr_Pcu_efficiencies_hp_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-14\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 4 ;// Number of poles in Induction motor\n",
+"f = 60 ; // Frequency in Hz\n",
+"V = 220 ; // Rated voltage of IM in volt\n",
+"hp_IM = 5 ; // Power rating of IM in hp\n",
+"PF = 0.9 ; // Power factor\n",
+"I_L = 16 ; // Line current in A\n",
+"S = 1750 ; // Speed of IM in rpm\n",
+"\n",
+"// No-load test data\n",
+"I_nl = 6.5 ; // No-load line current in A\n",
+"V_nl = 220 ; // No-load line voltage in volt\n",
+"P_nl = 300 ; // No-load power reading in W\n",
+"\n",
+"// Blocked rotor test\n",
+"I_br = 16 ; // Blocked rotor line current in A\n",
+"V_br = 50 ; // Blocked rotor voltage in volt\n",
+"P_br = 800 ; // Blocked rotor power reading in W\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"P_cu = P_br ; // Full-load equivalent cu-loss\n",
+"I_1 = I_br ; // Primary current in A\n",
+"R_e1 = (P_cu) / (3/2 * (I_1)^2 ); // Equivalent total resistance of IM in ohm \n",
+"\n",
+"// case b\n",
+"P_in = P_nl ; // Input power to IM\n",
+"I1 = I_nl ; // Input current in A\n",
+"P_r = P_in - (3/2 * (I1)^2 * R_e1); // Rotational losses in W\n",
+"\n",
+"// case c\n",
+"LF1 = 1/4 ; // Load fraction\n",
+"LF2 = 1/2 ; // Load fraction\n",
+"LF3 = 3/4 ; // Load fraction\n",
+"LF4 = 5/4 ; // Load fraction\n",
+"P_cu_LF1 = (LF1)^2 * P_cu ; // Equivalent copper loss at 1/4 rated-load\n",
+"P_cu_LF2 = (LF2)^2 * P_cu ; // Equivalent copper loss at 1/2 rated-load\n",
+"P_cu_LF3 = (LF3)^2 * P_cu ; // Equivalent copper loss at 3/4 rated-load\n",
+"P_cu_LF4 = (LF4)^2 * P_cu ; // Equivalent copper loss at 5/4 rated-load\n",
+"\n",
+"// case d\n",
+"Full_load_input = sqrt(3)*V*I_L*PF ; \n",
+"\n",
+"// Efficiency\n",
+"// Efficiency at 1/4 rated load\n",
+"eta_LF1 = ( Full_load_input*LF1 - (P_r + P_cu_LF1) ) / (Full_load_input*LF1) * 100 ;\n",
+"\n",
+"// Efficiency at 1/2 rated load\n",
+"eta_LF2 = ( Full_load_input*LF2 - (P_r + P_cu_LF2) ) / (Full_load_input*LF2) * 100 ;\n",
+"\n",
+"// Efficiency at 3/4 rated load\n",
+"eta_LF3 = ( Full_load_input*LF3 - (P_r + P_cu_LF3) ) / (Full_load_input*LF3) * 100 ;\n",
+"\n",
+"// Efficiency at rated load\n",
+"eta_rated = ( Full_load_input - (P_r + P_cu) ) / (Full_load_input) * 100 ;\n",
+"\n",
+"// Efficiency at 5/4 rated load\n",
+"eta_LF4 = ( Full_load_input*LF4 - (P_r + P_cu_LF4) ) / (Full_load_input*LF4) * 100 ;\n",
+"\n",
+"// case e\n",
+"// since eta is calculated in percent divide it by 100 for hp calculations\n",
+"P_o_LF1 = (Full_load_input*LF1*eta_LF1/100)/746 ; // Output hp at 1/4 rated load\n",
+"P_o_LF2 = (Full_load_input*LF2*eta_LF2/100)/746 ; // Output hp at 1/2 rated load\n",
+"P_o_LF3 = (Full_load_input*LF3*eta_LF3/100)/746 ; // Output hp at 3/4 rated load\n",
+"P_o = (Full_load_input*eta_rated/100)/746 ; // Output hp at 1/4 rated load\n",
+"P_o_LF4 = (Full_load_input*LF4*eta_LF4/100)/746 ; // Output hp at 5/4 rated load\n",
+"\n",
+"// case f\n",
+"hp = P_o ; // Rated output horsepower\n",
+"T_o = (P_o*5252)/S ; // Outpue torque at full-load in lb-ft\n",
+"T_o_Nm = T_o * 1.356 ; // Outpue torque at full-load in N-m\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-14 Solution : ');\n",
+"\n",
+"printf(' \n a: Equivalent total resistance of IM :\n    R_e1 = %.3f Ω \n',R_e1);\n",
+"\n",
+"printf(' \n b: Rotational losses :\n    P_r = %.f W \n ',P_r);\n",
+"\n",
+"printf(' \n c: At full-load, P_cu = %d W \n',P_cu);\n",
+"printf(' \n    P_cu at %.2f rated load = %d W',LF1,P_cu_LF1)\n",
+"printf(' \n    P_cu at %.2f rated load = %d W',LF2,P_cu_LF2)\n",
+"printf(' \n    P_cu at %.2f rated load = %d W',LF3,P_cu_LF3)\n",
+"printf(' \n    P_cu at %.2f rated load = %d W \n',LF4,P_cu_LF4)\n",
+"\n",
+"printf(' \n d: Full-load input = %.f W \n',Full_load_input);\n",
+"printf(' \n    Efficiency :\n    η at %.2f rated load = %.1f percent \n',LF1,eta_LF1);\n",
+"printf(' \n    η at %.2f rated load = %.1f percent \n',LF2,eta_LF2);\n",
+"printf(' \n    η at %.2f rated load = %.1f percent \n',LF3,eta_LF3);\n",
+"printf(' \n    η at rated load = %.1f percent \n',eta_rated);\n",
+"printf(' \n    η at %.2f rated load = %.1f percent \n',LF4,eta_LF4);\n",
+"\n",
+"printf(' \n e: Output horsepower :\n    P_o at %.2f rated load = %.3f hp \n',LF1,P_o_LF1);\n",
+"printf(' \n    P_o at %.2f rated load = %.3f hp \n',LF2,P_o_LF2);\n",
+"printf(' \n    P_o at %.2f rated load = %.3f hp \n',LF3,P_o_LF3);\n",
+"printf(' \n    P_o at rated load = %.3f hp \n',P_o);\n",
+"printf(' \n    P_o at %.2f rated load = %.3f hp \n',LF4,P_o_LF4);\n",
+"\n",
+"printf(' \n f: Output torque at full-load :\n    T_o = %.1f lb-ft',T_o);\n",
+"printf(' \n    T_o = %.2f N-m',T_o_Nm);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.15: RPO_efficiency_hp_torque_compare.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-15\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data(from Ex.12-14)\n",
+"pole = 4 ;// Number of poles in Induction motor\n",
+"f = 60 ; // Frequency in Hz\n",
+"V = 220 ; // Rated voltage of IM in volt\n",
+"hp_IM = 5 ; // Power rating of IM in hp\n",
+"PF = 0.9 ; // Power factor\n",
+"I_L = 16 ; // Line current in A\n",
+"S_r = 1750 ; // Speed of IM in rpm\n",
+"\n",
+"// No-load test data\n",
+"I_nl = 6.5 ; // No-load line current in A\n",
+"V_nl = 220 ; // No-load line voltage in volt\n",
+"P_nl = 300 ; // No-load power reading in W\n",
+"\n",
+"// Blocked rotor test\n",
+"I_br = 16 ; // Blocked rotor line current in A\n",
+"V_br = 50 ; // Blocked rotor voltage in volt\n",
+"P_br = 800 ; // Blocked rotor power reading in W\n",
+"R_dc = 1 ; // dc resistance in ohm between lines \n",
+"\n",
+"// given data from ex.12-15\n",
+"V = 220 ; // voltage rating in volt\n",
+"P_input = 5500 ; // power drawn in W\n",
+"\n",
+"// Calculations\n",
+"// Preliminary calculations\n",
+"R_e1 = 1.25*R_dc ; // Equivalent total resistance of IM in ohm\n",
+"P_in = P_nl ; // Input power to IM in W\n",
+"I1 = I_nl ; // Input current in A\n",
+"P_r = P_in - (3/2 * (I1)^2 * R_e1); // Rotational losses in W\n",
+"\n",
+"I_1 = I_L ;\n",
+"SCL = (3/2 * (I_1)^2 * R_e1) ; // Stator Copper Loss in W at full-load\n",
+"SPI = P_input ; // Stator Power Input in W\n",
+"RPI = SPI - SCL ; // Rotor Power Input in W\n",
+"\n",
+"S = (120*f/pole); // Speed of synchronous magnetic field in rpm\n",
+"s = (S-S_r)/S ; // Slip\n",
+"\n",
+"RPD = RPI*(1-s); // Rotor Power Developed in W\n",
+"RPO = RPD - P_r ; // Rotor Power Output in W\n",
+"\n",
+"// case a\n",
+"P_o = RPO ;\n",
+"eta_fl = (P_o / P_input)*100 ; // Full-load efficiency\n",
+"\n",
+"// case b\n",
+"hp = P_o / 746 ; // Output horsepower\n",
+"T_o = (hp*5252)/S_r ; // Output torque in lb-ft\n",
+"T_o_Nm = T_o * 1.356 ; // Output torque in N-m\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-15 Solution : '); \n",
+"\n",
+"printf(' \n Preliminary calculations :');\n",
+"printf(' \n R_e1 = %.2f Ω \n',R_e1);\n",
+"printf(' \n P_r = %.1f W \n ',P_r);\n",
+"printf(' \n SCL(fl) = %d W \n ',SCL);\n",
+"printf(' \n RPI(fl) = %d W \n ',RPI);\n",
+"printf(' \n RPD(fl) = %f W ≃ %.1f W \n ',RPD,RPD);\n",
+"printf(' \n RPO(fl) = %f W ≃ %.f W \n ',RPO,RPO);\n",
+"\n",
+"printf(' \n a: Full-load efficiency :\n    η_fl = %.1f percent \n',eta_fl);\n",
+"\n",
+"printf(' \n b: Output horsepower :\n    hp = %.2f hp at full-load \n',hp);\n",
+"printf(' \n    Output torque at full-load :\n    T_o = %f lb-ft ≃ %.1f lb-ft',T_o,T_o);\n",
+"printf(' \n    T_o = %f lb-ft ≃ %.2f N-m \n ',T_o_Nm,T_o_Nm);\n",
+"\n",
+"printf(' \n c: Comparision of results');\n",
+"printf(' \n ________________________________________________________________');\n",
+"printf(' \n \t\t\t\t\t Ex.12-14\tEx.12-15');\n",
+"printf(' \n ________________________________________________________________');\n",
+"printf(' \n \t η_fl(percent) \t\t\t 82.4 \t\t %.1f ',eta_fl);\n",
+"printf(' \n \t Rated output(hp) \t\t 6.06 \t\t %.2f ',hp);\n",
+"printf(' \n \t Rated output torque(lb-ft) \t 18.2 \t\t %.1f ',T_o);\n",
+"printf(' \n ________________________________________________________________');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.16: Ip_Ir_PF_SPI_SCL_RPI_RCL_RPD_T_hp_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-16\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// code letter = J\n",
+"P = 6 ; // Number of poles\n",
+"S_r = 1176 ; // rotor speed in rpm\n",
+"V = 220 ; // Rated voltage of SCIM in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"hp_SCIM = 7.5 ; // Power rating of SCIM in hp\n",
+"\n",
+"R_ap = 0.3 ; // armature resistance in ohm/phase\n",
+"R_r = 0.144 ; // rotor resistance in ohm/phase\n",
+"jX_m = 13.5 ; // reactance in ohm/phase\n",
+"jX_s = 0.5 ; // synchronous reactance in ohm/phase\n",
+"jX_lr = 0.2 ; // Locked rotor reactance in ohm/phase\n",
+"P_r = 300 ; // Rotational losses in W\n",
+"\n",
+"disp('Example 12-16 : '); \n",
+"// Calculations\n",
+"S = (120*f/P); // Speed of synchronous magnetic field in rpm\n",
+"// case a\n",
+"s = (S-S_r)/S ; // Slip\n",
+"\n",
+"R_r_by_s = R_r / s ;\n",
+"\n",
+"// case b\n",
+"printf(' \n From fig.12-11 , using the format method of mesh analysis,we may write');\n",
+"printf(' \n the array by inspection :\n ');\n",
+"printf(' \n_______________________________________________');\n",
+"printf(' \n \t     I_1 \t I_2 \t\t V ');\n",
+"printf(' \n_______________________________________________');\n",
+"printf(' \n\t (0.3+j14)   -(0+j13.5) \t(127+j0)');\n",
+"printf(' \n\t-(0+j13.5)   (7.2+j13.7) \t 0');\n",
+"printf(' \n_______________________________________________\n');\n",
+"\n",
+"A = [ (0.3 + %i*14) -%i*13.5 ; (-%i*13.5)  (7.2 + %i*13.7) ]; // Matrix containing above mesh eqns array\n",
+"delta = det(A); // Determinant of A\n",
+"\n",
+"// case b : Stator armature current I_p in A\n",
+"I_p = det( [ (127+%i*0) (-%i*13.5) ; 0 (7.2 + %i*13.7) ] ) / delta ;\n",
+"I_p_m = abs(I_p);//I_p_m=magnitude of I_p in A\n",
+"I_p_a = atan(imag(I_p) /real(I_p))*180/%pi;//I_p_a=phase angle of I_p in degrees\n",
+"I_1 = I_p ; // Stator armature current in A\n",
+"\n",
+"// case c : Rotor current I_r per phase in A\n",
+"I_r = det( [ (0.3 + %i*14) (127+%i*0) ; (-%i*13.5) 0 ] ) / delta ;\n",
+"I_r_m = abs(I_r);//I_r_m=magnitude of I_r in A\n",
+"I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees\n",
+"\n",
+"// case d\n",
+"theta = I_p_a ; // Motor PF angle in degrees\n",
+"cos_theta = cosd(theta); // Motor PF\n",
+"\n",
+"// case e\n",
+"I_p = I_p_m ; // Stator armature current in A\n",
+"V_p = V / sqrt(3); // Phase voltage in volt\n",
+"SPI = V_p * I_p * cos_theta ; // Stator Power Input in W\n",
+"\n",
+"// case f\n",
+"SCL = (I_p)^2 * R_ap ; // Stator Copper Loss in W\n",
+"\n",
+"// case g\n",
+"// Subscripts 1 and 2 for RPI indicates two methods of calculating RPI\n",
+"RPI_1 = SPI - SCL ; // Rotor Power Input in W\n",
+"RPI_2 = (I_r_m)^2 * (R_r/s); // Rotor Power Input in W\n",
+"RPI =RPI_1 ;\n",
+"\n",
+"// case h\n",
+"RCL = s*(RPI); // Rotor copper losses in W\n",
+"\n",
+"// case i\n",
+"// Subscripts 1 , 2 and 3 for RPD indicates three methods of calculating RPD\n",
+"RPD_1 = RPI - RCL ; // Rotor Power Developed in W\n",
+"RPD_2 = RPI * ( 1 - s ); // Rotor Power Developed in W\n",
+"RPD = RPD_1 ;\n",
+"\n",
+"// case j\n",
+"RPO = 3*RPD - P_r ; // Rotor Power Developed in W\n",
+"\n",
+"// case k\n",
+"P_to = RPO ; // Total rotor power in W\n",
+"T_o = (7.04*P_to)/S_r ; // Total 3-phase torque in lb-ft\n",
+"\n",
+"// case l\n",
+"hp = P_to / 746 ; // Output horsepower\n",
+"\n",
+"// case m\n",
+"P_in = 3*SPI ; // Input power to stator in W\n",
+"P_o = RPO ; // Output power in W\n",
+"eta = P_o / P_in * 100 ; // Motor efficiency at rated load\n",
+"\n",
+"// Display the results\n",
+"disp('Solution : '); \n",
+"printf(' \n a: s = %.2f \n    R_r/s = %.1f Ω \n',s,R_r_by_s );\n",
+"\n",
+"printf(' \n    Determinant Δ = ');disp(delta);\n",
+"\n",
+"printf(' \n b: Stator armature current :\n    I_p in A = ');disp(I_1);\n",
+"printf(' \n    I_p = I_1 = %.2f <%.2f A \n ',I_p_m , I_p_a );\n",
+"\n",
+"printf(' \n c: Rotor current per phase :\n    I_r in A = ');disp(I_r);\n",
+"printf(' \n    I_r = I_2 = %.3f <%.2f A \n ',I_r_m , I_r_a );\n",
+"\n",
+"printf(' \n d: Motor PF :\n    cosӨ = %.4f \n',cos_theta);\n",
+"\n",
+"printf(' \n e: Stator Power Input :\n    SPI = %d W \n',SPI);\n",
+"\n",
+"printf(' \n f: Stator Copper Loss :\n    SCL = %.1f W \n',SCL);\n",
+"\n",
+"printf(' \n g: Rotor Power Input :\n    RPI = %.1f W(method 1) ', RPI_1);\n",
+"printf(' \n    RPI = %.1f W (method 2)\n',RPI_2);\n",
+"\n",
+"printf(' \n h: Rotor copper loss :\n    RCL = %.1f W\n',RCL);\n",
+"\n",
+"printf(' \n i: Rotor Power Developed :\n    RPD = %.1f W \n',RPD_1);\n",
+"\n",
+"printf(' \n    RPD = %.1f W \n ',RPD_2);\n",
+"\n",
+"printf(' \n j: Total 3-phase rotor power:\n    RPO = %f W \n',RPO);\n",
+"\n",
+"printf(' \n k: Total output torque developed :\n    T_o = %.2f lb-ft\n',T_o);\n",
+"\n",
+"printf(' \n l: Output horsepower : \n    hp = %.2f hp (rated 7.5 hp)\n',hp);\n",
+"\n",
+"printf(' \n m: Motor efficiency at rated load :\n    η = %.2f percent \n',eta);\n",
+"\n",
+"printf(' \n n: See Fig.12-12');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.17: upper_and_lower_limit_Is.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-17\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// code letter = J of SCIM (Ex.12-16)\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// From Appendix A-3,Table 430-7(b),the starting kVA/hp (with rotor locked) is \n",
+"// less than 7.99,which, when substituted in the following equation, yields a\n",
+"// maximum starting current of :\n",
+"\n",
+"// subscript u for I_s indicates upper limit of starting current\n",
+"I_s_u = (7.99*(7.5*1000))/(sqrt(3)*220) ;\n",
+"\n",
+"// case b\n",
+"// The lower limit,code letter J,is 7.1 kVA/hp. Thus :\n",
+"\n",
+"// subscript l for I_s indicates lower limit of starting current\n",
+"I_s_l = (7.1*(7.5*1000))/(sqrt(3)*220) ;\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-17 Solution : ');\n",
+"\n",
+"printf(' \n a: From Appendix A-3,Table 430-7(b),the starting kVA/hp ');\n",
+"printf(' \n    (with rotor locked) is less than 7.99,which, when substituted ');\n",
+"printf(' \n    in the following equation, yields a maximum starting current of :');\n",
+"printf(' \n    I_s = %.1f A \n',I_s_u);\n",
+"\n",
+"printf(' \n b: The lower limit,code letter J,is 7.1 kVA/hp.\n    Thus :');\n",
+"printf(' \n    I_s = %.1f A ',I_s_l );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.18: starting_I_and_PF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-18\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data (Ex.12-16)\n",
+"// code letter = J\n",
+"P = 6 ; // Number of poles\n",
+"S_r = 1176 ; // rotor speed in rpm\n",
+"V = 220 ; // Rated voltage of SCIM in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"hp_SCIM = 7.5 ; // Power rating of SCIM in hp\n",
+"\n",
+"R_ap = 0.3 ; // armature resistance in ohm/phase\n",
+"R_r = 0.144 ; // rotor resistance in ohm/phase\n",
+"jX_m = 13.5 ; // reactance in ohm/phase\n",
+"jX_s = 0.5 ; // synchronous reactance in ohm/phase\n",
+"jX_lr = 0.2 ; // Locked rotor reactance in ohm/phase\n",
+"P_r = 300 ; // Rotational losses in W\n",
+"s = 1 ; // unity slip\n",
+"\n",
+"disp('Example 12-18 Solution : ');\n",
+"\n",
+"printf(' \n The ratio R_r/s = %.3f ohm, in fig.12-11 , using the format method ',R_r/s);\n",
+"printf(' \n of mesh analysis,we may write the array by inspection :\n ');\n",
+"printf(' \n_______________________________________________');\n",
+"printf(' \n \t     I_1 \t I_2 \t\t V ');\n",
+"printf(' \n_______________________________________________');\n",
+"printf(' \n\t (0.3+j14)   -(0+j13.5) \t(127+j0)');\n",
+"printf(' \n\t-(0+j13.5)   (0.144+j13.7) \t 0');\n",
+"printf(' \n_______________________________________________\n');\n",
+"\n",
+"// Calculations\n",
+"\n",
+"A = [ (0.3 + %i*14) -%i*13.5 ; (-%i*13.5)  (0.144 + %i*13.7) ]; // Matrix containing above mesh eqns array\n",
+"delta = det(A); // Determinant of A\n",
+"\n",
+"// case a : Starting stator current I_s per phase in A\n",
+"I_s = det( [ (127+%i*0) (-%i*13.5) ; 0 (0.144 + %i*13.7) ] ) / delta ;\n",
+"I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A\n",
+"I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees\n",
+"\n",
+"// case b : power factor of the motor at starting\n",
+"theta = I_s_a ; // Motor PF angle in degrees\n",
+"cos_theta = cosd(theta); // Motor PF\n",
+"\n",
+"// Display the results\n",
+"disp('Solution : '); \n",
+"printf(' \n a: Starting stator current of SCIM :\n    I_s = I_1 = ');disp(I_s);\n",
+"printf(' \n    I_s = I_1 = %.2f <%.2f A \n ',I_s_m , I_s_a );\n",
+"\n",
+"printf(' \n b: Power factor of the motor at starting :\n    cosӨ = %.4f ≃ %.3f\n',cos_theta,cos_theta);\n",
+"\n",
+"printf(' \n    Note : I_s = %.2f A calculated in Ex.12-18 falls between the limits',I_s_m);\n",
+"printf(' \n           found in Ex.12-17. This verifies the mesh analysis technique.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.19: Re1s_slip_Pcu_and_Pr_at_LFs_hp_T.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-19\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 220 ; // Rated voltage of SCIM in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"P = 4 ; // Number of poles\n",
+"PF = 0.85 ; // power factor of capacitor start IM\n",
+"// nameplate details\n",
+"hp_IM = 5 ; // power rating of IM in hp\n",
+"I_L = 28 ; // Rated line current in A\n",
+"S_r = 1620 ; // Rotor speed of IM in rpm\n",
+"\n",
+"// No-load test data\n",
+"I_nl = 6.4 ; // No-load line current in A\n",
+"V_nl = 220 ; // No-load line voltage in volt\n",
+"P_nl = 239 ; // No-load power reading in W\n",
+"s_nl = 0.01 ; // No-load slip\n",
+"\n",
+"// Blocked rotor test\n",
+"I_br = 62 ; // Blocked rotor line current in A\n",
+"V_br = 64 ; // Blocked rotor voltage in volt\n",
+"P_br = 1922 ; // Blocked rotor power reading in W\n",
+"s_br = 1 ; // blocked rotor slip(unity) \n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"R_e1s = P_br / (I_br^2); // Equivalent total resistance of IM in ohm\n",
+"\n",
+"// case b\n",
+"P_in = P_nl ; // Input power to IM in W\n",
+"I_1s = I_nl ; // Input current in A\n",
+"P_ro = P_in - ((I_1s)^2 * R_e1s); // Rotational losses in W\n",
+"\n",
+"// case c\n",
+"S = (120*f/P); // Speed of synchronous magnetic field in rpm\n",
+"S_fl = S_r ; // Full-load rotor speed of IM in rpm \n",
+"s_fl = (S - S_fl)/S ; // Full-load Slip \n",
+"\n",
+"LF1 = 1/4 ; // Load fraction\n",
+"LF2 = 1/2 ; // Load fraction\n",
+"LF3 = 3/4 ; // Load fraction\n",
+"LF4 = 5/4 ; // Load fraction\n",
+"\n",
+"s_LF1 = s_fl*LF1 ; // slip at 1/4 rated load\n",
+"s_LF2 = s_fl*LF2 ; // slip at 1/2 rated load\n",
+"s_LF3 = s_fl*LF3 ; // slip at 3/4 rated load\n",
+"s_LF4 = s_fl*LF4 ; // slip at 5/4 rated load\n",
+"\n",
+"// case d\n",
+"s_o = s_nl ; // No-load slip\n",
+"P_rs_LF1 = P_ro * (1 - s_LF1)/(1 - s_o); // Rotational losses in W at s_LF1\n",
+"P_rs_LF2 = P_ro * (1 - s_LF2)/(1 - s_o); // Rotational losses in W at s_LF2\n",
+"P_rs_LF3 = P_ro * (1 - s_LF3)/(1 - s_o); // Rotational losses in W at s_LF3\n",
+"P_rs_fl = P_ro * (1 - s_fl)/(1 - s_o); // Rotational losses in W at full-load slip\n",
+"P_rs_LF4 = P_ro * (1 - s_LF4)/(1 - s_o); // Rotational losses in W at s_LF4\n",
+"\n",
+"// case e\n",
+"I1s = I_L ; //  Line current in A\n",
+"P_cu_fl = (I1s)^2*R_e1s ; // Equivalent copper loss at full-load slip\n",
+"P_cu_LF1 = (LF1)^2 * P_cu_fl ; // Equivalent copper loss at s_LF1\n",
+"P_cu_LF2 = (LF2)^2 * P_cu_fl ; // Equivalent copper loss at s_LF2\n",
+"P_cu_LF3 = (LF3)^2 * P_cu_fl ; // Equivalent copper loss at s_LF3\n",
+"P_cu_LF4 = (LF4)^2 * P_cu_fl ; // Equivalent copper loss at s_LF4\n",
+"\n",
+"// case f\n",
+"Input = V*I_L*PF ; // Input to single phase capacitor start IM\n",
+"\n",
+"// Efficiency at 1/4 rated load\n",
+"eta_LF1 = ( Input*LF1 - (P_rs_LF1 + P_cu_LF1) ) / (Input*LF1) * 100 ;\n",
+"\n",
+"// Efficiency at 1/2 rated load\n",
+"eta_LF2 = ( Input*LF2 - (P_rs_LF2 + P_cu_LF2) ) / (Input*LF2) * 100 ;\n",
+"\n",
+"// Efficiency at 3/4 rated load\n",
+"eta_LF3 = ( Input*LF3 - (P_rs_LF3 + P_cu_LF3) ) / (Input*LF3) * 100 ;\n",
+"\n",
+"// Efficiency at rated load\n",
+"eta_fl = ( Input - (P_rs_fl + P_cu_fl) ) / (Input) * 100 ;\n",
+"\n",
+"// Efficiency at 5/4 rated load\n",
+"eta_LF4 = ( Input*LF4 - (P_rs_LF4 + P_cu_LF4) ) / (Input*LF4) * 100 ;\n",
+"\n",
+"// case g\n",
+"// since eta is calculated in percent divide it by 100 for hp calculations\n",
+"P_o_LF1 = (Input*LF1*eta_LF1/100)/746 ; // Output hp at 1/4 rated load\n",
+"P_o_LF2 = (Input*LF2*eta_LF2/100)/746 ; // Output hp at 1/2 rated load\n",
+"P_o_LF3 = (Input*LF3*eta_LF3/100)/746 ; // Output hp at 3/4 rated load\n",
+"P_o = (Input*eta_fl/100)/746 ; // Output hp at 1/4 rated load\n",
+"P_o_LF4 = (Input*LF4*eta_LF4/100)/746 ; // Output hp at 5/4 rated load\n",
+"\n",
+"// case h\n",
+"hp = P_o ; // Rated output horsepower\n",
+"S_fl = S_r ; // Full-load rotor speed in rpm\n",
+"T_o = (P_o*5252)/S_fl ; // Outpue torque at full-load in lb-ft\n",
+"T_o_Nm = T_o * 1.356 ; // Outpue torque at full-load in N-m\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-19 Solution : '); \n",
+"\n",
+"printf(' \n a: Equivalent total resistance of IM :\n    R_e1s = %.1f Ω \n',R_e1s);\n",
+"\n",
+"printf(' \n b: Rotational losses :\n    P_ro = %.1f W \n ',P_ro);\n",
+"\n",
+"printf(' \n c: Slip at rated load : s = %.1f \n    Slip,',s_fl);\n",
+"printf(' \n    s at %.2f rated load = %.3f',LF1,s_LF1);\n",
+"printf(' \n    s at %.2f rated load = %.3f',LF2,s_LF2);\n",
+"printf(' \n    s at %.2f rated load = %.3f',LF3,s_LF3);\n",
+"printf(' \n    s at %.2f rated load = %.3f \n ',LF4,s_LF4);\n",
+"\n",
+"printf(' \n d: Rotational losses :\n ');\n",
+"printf(' \n    P_r at at %.2f rated load = %.1f W ',LF1,P_rs_LF1);\n",
+"printf(' \n    P_r at at %.2f rated load = %.1f W ',LF2,P_rs_LF2);\n",
+"printf(' \n    P_r at at %.2f rated load = %.1f W ',LF3,P_rs_LF3);\n",
+"printf(' \n    P_r at at full load = %.1f W ',P_rs_fl);\n",
+"printf(' \n    P_r at at %.2f rated load = %.1f W \n ',LF4,P_rs_LF4);\n",
+"\n",
+"printf(' \n e: At full-load, P_cu = %d W \n',P_cu_fl);\n",
+"printf(' \n    P_cu at %.2f rated load = %.2f W',LF1,P_cu_LF1)\n",
+"printf(' \n    P_cu at %.2f rated load = %.2f W',LF2,P_cu_LF2)\n",
+"printf(' \n    P_cu at %.2f rated load = %.2f W',LF3,P_cu_LF3)\n",
+"printf(' \n    P_cu at %.2f rated load = %.2f W \n',LF4,P_cu_LF4)\n",
+"\n",
+"printf(' \n f: Full-load input = %.f W \n',Input);\n",
+"printf(' \n    Efficiency :\n    η at %.2f rated load = %.1f percent \n',LF1,eta_LF1);\n",
+"printf(' \n    η at %.2f rated load = %.1f percent \n',LF2,eta_LF2);\n",
+"printf(' \n    η at %.2f rated load = %.1f percent \n',LF3,eta_LF3);\n",
+"printf(' \n    η at rated load = η_fl = %.1f percent \n',eta_fl);\n",
+"printf(' \n    η at %.2f rated load = %.1f percent \n',LF4,eta_LF4);\n",
+"printf(' \n    Please note: Calculation error for η_fl in textbook.\n');\n",
+"\n",
+"printf(' \n g: Output horsepower :\n    P_o at %.2f rated load = %.3f hp \n',LF1,P_o_LF1);\n",
+"printf(' \n    P_o at %.2f rated load = %.3f hp \n',LF2,P_o_LF2);\n",
+"printf(' \n    P_o at %.2f rated load = %.3f hp \n',LF3,P_o_LF3);\n",
+"printf(' \n    P_o at rated load = %.3f hp \n',P_o);\n",
+"printf(' \n    P_o at %.2f rated load = %.3f hp \n',LF4,P_o_LF4);\n",
+"\n",
+"printf(' \n h: Output torque at full-load :\n    T_o = %.1f lb-ft',T_o);\n",
+"printf(' \n    T_o = %.2f N-m ≃ %.1f N-m',T_o_Nm,T_o_Nm);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.1: Pr_Ia_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 10000 ; // Power rating of the shunt generator in W\n",
+"V = 230 ;// Voltage rating of the shunt generator in volt\n",
+"S = 1750 ; // Speed in rpm of the shunt generator\n",
+"// Shunt generator was made to run as a motor\n",
+"V_a = 245 ; // Voltage across armature in volt\n",
+"I_a = 2 ; // Armature current in A\n",
+"R_f = 230 ; // Field resistance in ohm\n",
+"R_a = 0.2 ; // Armature resistance\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"Rotational_losses = (V_a * I_a) - (I_a^2 * R_a); // Rotational losses in W at full load \n",
+"\n",
+"// case b\n",
+"V_t = V ;\n",
+"// At rated load\n",
+"I_L = P / V_t ; // Line current in A\n",
+"I_f = V / R_f ; // Field current in A\n",
+"Ia = I_f + I_L ; // Armature current in A\n",
+"\n",
+"armature_loss = (Ia^2 * R_a); // Full-load armature loss in W\n",
+"V_f = V ; // Field voltage in volt\n",
+"field_loss = V_f * I_f; // Full-load field loss in W\n",
+"\n",
+"// case c\n",
+"//\n",
+"eta = P / ( P + Rotational_losses + (armature_loss+field_loss) ) * 100 ; \n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-1 Solution : ');\n",
+"\n",
+"printf(' \n a: Rotational losses at full load = %.1f W \n',Rotational_losses);\n",
+"\n",
+"printf(' \n b: At the rated load,\n    I_L = %.1f A\n    I_a = %.1f A\n',I_L,Ia);\n",
+"printf(' \n    Full-load armature loss :\n    (I_a^2)*R_a = %.f W \n',armature_loss);\n",
+"printf(' \n    Full-load field loss :\n    V_f*I_f = %.f W \n',field_loss);\n",
+"\n",
+"printf(' \n c: Efficiency of the generator at rated load(full-load in this Ex.) : ');\n",
+"printf(' \n    η = %.1f percent ',eta);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.2: efficiency_at_different_LF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// data from Ex.12-1\n",
+"P = 10000 ; // Power rating of the shunt generator in W\n",
+"V = 230 ;// Voltage rating of the shunt generator in volt\n",
+"S = 1750 ; // Speed in rpm of the shunt generator\n",
+"\n",
+"// ( Solutions from Example 12-1 )\n",
+"Rotational_losses = 489.2 // Rotational losses at full load in W\n",
+"armature_loss = 396 ; // Full-load armature loss in W\n",
+"field_loss = 230 ; // Full-load field loss in W\n",
+"\n",
+"// case a\n",
+"x1 = (1/4); // Fraction of full-load\n",
+"// Subscript a for eta indicates case a\n",
+"eta_a = (P*x1) / ( (P*x1) + Rotational_losses + (armature_loss*(x1^2)+field_loss) ) * 100 ; \n",
+"\n",
+"// case b\n",
+"x2 = (1/2); // Fraction of full-load\n",
+"// Subscript b for eta indicates case b\n",
+"eta_b = (P*x2) / ( (P*x2) + Rotational_losses + (armature_loss*(x2^2)+field_loss) ) * 100 ; \n",
+"\n",
+"// case c\n",
+"x3 = (3/4); // Fraction of full-load\n",
+"// Subscript c for eta indicates case c\n",
+"eta_c = (P*x3) / ( (P*x3) + Rotational_losses + (armature_loss*(x3^2)+field_loss) ) * 100 ; \n",
+"\n",
+"// case d\n",
+"x4 = (5/4); // Fraction of full-load\n",
+"// Subscript d for eta indicates case d\n",
+"eta_d = (P*x4) / ( (P*x4) + Rotational_losses + (armature_loss*(x4^2)+field_loss) ) * 100 ; \n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-2 Solution : ');\n",
+"\n",
+"printf(' \n    If x is the fraction of full-load, then \n ');\n",
+"printf(' \n a: Efficiency of generator when x = %.2f ',x1 );\n",
+"printf(' \n    η = %.1f percent \n ',eta_a);\n",
+"\n",
+"printf(' \n b: Efficiency of generator when x = %.2f ',x2 );\n",
+"printf(' \n    η = %.1f percent \n ',eta_b);\n",
+"\n",
+"printf(' \n c: Efficiency of generator when x = %.2f ',x3 );\n",
+"printf(' \n    η = %.1f percent \n ',eta_c);\n",
+"\n",
+"printf(' \n d: Efficiency of generator when x = %.2f ',x4 );\n",
+"printf(' \n    η = %.1f percent \n ',eta_d);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.3: field_current_Ec_Pf.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 240 ; // Voltage rating of the dc shunt motor in volt\n",
+"P_hp = 25 ; // Power rating of the dc shunt motor in hp\n",
+"S = 1800 ; // Speed in rpm of the shunt generator\n",
+"I_L = 89 ; // Full-load line current\n",
+"R_a = 0.05 ; // Armature resistance in ohm\n",
+"R_f = 120 ; // Field resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_f = V ; // Field voltage in volt\n",
+"I_f = V_f / R_f ; // Field current in A\n",
+"I_a = I_L - I_f  ; // Armature current in A\n",
+"V_a = V ;\n",
+"E_c = V_a - I_a*R_a ; // Armature voltage to be applied to the motor when motor \n",
+"// is run light at 1800 rpm during stray power test\n",
+"\n",
+"// case b\n",
+"Ia = 4.2 ; // Armature current in A produced by E_c\n",
+"Va = E_c ; // Armature voltage in volt\n",
+"P_r = Va*Ia ; // Stray power in W ,when E_c produces I_a = 4.2 A at speed of 1800 rpm\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-3 Solution : ');\n",
+"\n",
+"printf(' \n a: Field current :\n    I_f = %d A \n ',I_f );\n",
+"printf(' \n    Armature current :\n    I_a = %d A \n ',I_a );\n",
+"printf(' \n    Armature voltage to be applied to the motor when motor is run');\n",
+"printf(' \n    light at %d rpm during stray power test :\n ',S );\n",
+"printf(' \n    E_c = %.2f V \n ',E_c );\n",
+"\n",
+"printf(' \n b: Stray power :\n    P_r = %.1f W ',P_r ); "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.4: Pr_variable_losses_efficiency_table.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 600 ; // Voltage rating of the compound motor in volt\n",
+"P_hp = 150 ; // Power rating of the compound motor in hp\n",
+"I_L = 205 ; // Full-load rated line current in A\n",
+"S = 1500 ; // Full-load Speed in rpm of the compound generator\n",
+"R_sh = 300 ; // Shunt field resistance in ohm\n",
+"R_a = 0.05 ; // Armature resistance in ohm\n",
+"R_s = 0.1 ; // Series field resistance in ohm\n",
+"V_a  = 570 ; // Applied voltage in volt\n",
+"I_a = 6 ; // Armature current in A\n",
+"S_o = 1800 ; // No-load Speed in rpm of the compound generator\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"Rot_losses = V_a*I_a ; // Rotational losses in W\n",
+"// If x is fraction of full-load\n",
+"x1 = (1/4);\n",
+"S_1 = S_o - 300*x1 ; // Speed at 1/4 load\n",
+"Rot_losses_S_1 = (S_1/S)*Rot_losses ; // Rotational losses in W at speed S_1\n",
+"\n",
+"x2 = (1/2);\n",
+"S_2 = S_o - 300*x2 ; // Speed at 1/2 load\n",
+"Rot_losses_S_2 = (S_2/S)*Rot_losses ; // Rotational losses in W at speed S_2\n",
+"\n",
+"x3 = (3/4);\n",
+"S_3 = S_o - 300*x3 ; // Speed at 3/4 load\n",
+"Rot_losses_S_3 = (S_3/S)*Rot_losses ; // Rotational losses in W at speed S_3\n",
+"\n",
+"x4 = (5/4);\n",
+"S_4 = S_o - 300*x4 ; // Speed at 5/4 load\n",
+"Rot_losses_S_4 = (S_4/S)*Rot_losses ; // Rotational losses in W at speed S_4\n",
+"\n",
+"// case b\n",
+"I_sh = V / R_sh ; // Full-load shunt field current in A\n",
+"Ia = I_L - I_sh ; // Full-load armature current in A\n",
+"FL_variable_loss = (Ia^2)*(R_a + R_s); // Full-load variable electric losses in W\n",
+"\n",
+"x1_variable_loss = FL_variable_loss * (x1)^2 ; // Variable losses at 1/4 load\n",
+"x2_variable_loss = FL_variable_loss * (x2)^2 ; // Variable losses at 1/2 load\n",
+"x3_variable_loss = FL_variable_loss * (x3)^2 ; // Variable losses at 3/4 load\n",
+"x4_variable_loss = FL_variable_loss * (x4)^2 ; // Variable losses at 5/4 load\n",
+"\n",
+"// case c\n",
+"// Efficiency of motor = (Input - losses)/Input\n",
+"// where Input = volts*amperes*load_fraction\n",
+"//       Losses = field loss + rotational losses + variable electric losses\n",
+"// Input\n",
+"Input_FL = V * I_L ; // Input in W at full load\n",
+"Input_x1 = V * I_L * x1 ; // Input in W at 1/4 load\n",
+"Input_x2 = V * I_L * x2 ; // Input in W at 1/2 load\n",
+"Input_x3 = V * I_L * x3 ; // Input in W at 3/4 load\n",
+"Input_x4 = V * I_L * x4 ; // Input in W at 5/4 load\n",
+"\n",
+"Field_loss = V * I_sh // Field loss for each of the conditions of load\n",
+"\n",
+"// Rotational losses are calculated in part a while variable electric losses in part b\n",
+"\n",
+"// Total losses \n",
+"Losses_FL = Field_loss + Rot_losses + FL_variable_loss ; // Total losses for full load \n",
+"Losses_1 = Field_loss + Rot_losses_S_1 + x1_variable_loss ; // Total losses for 1/4 load \n",
+"Losses_2 = Field_loss + Rot_losses_S_2 + x2_variable_loss ; // Total losses for 1/2 load \n",
+"Losses_3 = Field_loss + Rot_losses_S_3 + x3_variable_loss ; // Total losses for 3/4 load \n",
+"Losses_4 = Field_loss + Rot_losses_S_4 + x4_variable_loss ; // Total losses for 5/4 load \n",
+"\n",
+"// Efficiency\n",
+"eta_FL = ( (Input_FL - Losses_FL) / Input_FL ) ; // Efficiency for 1/4 load\n",
+"eta_1 = ( (Input_x1 - Losses_1) / Input_x1 ) ; // Efficiency for 1/4 load \n",
+"eta_2 = ( (Input_x2 - Losses_2) / Input_x2 ) ; // Efficiency for 1/2 load \n",
+"eta_3 = ( (Input_x3 - Losses_3) / Input_x3 ) ; // Efficiency for 3/4 load \n",
+"eta_4 = ( (Input_x4 - Losses_4) / Input_x4 ) ; // Efficiency for 5/4 load \n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-4 Solution : ');\n",
+"\n",
+"printf(' \n a: Rotational loss = %d W at %d rpm(rated load)\n',Rot_losses,S);\n",
+"printf(' \n    Speed at %.2f load = %d rpm  ',x1 , S_1 );\n",
+"printf(' \n    Rotational loss at %d rpm = %d W \n ', S_1 , Rot_losses_S_1 );\n",
+"\n",
+"printf(' \n    Speed at %.2f load = %d rpm  ',x2 , S_2 );\n",
+"printf(' \n    Rotational loss at %d rpm = %d W \n ', S_2 , Rot_losses_S_2 );\n",
+"\n",
+"printf(' \n    Speed at %.2f load = %d rpm  ',x3 , S_3 );\n",
+"printf(' \n    Rotational loss at %d rpm = %d W \n ', S_3 , Rot_losses_S_3 );\n",
+"\n",
+"printf(' \n    Speed at %.2f load = %d rpm  ',x4 , S_4 );\n",
+"printf(' \n    Rotational loss at %d rpm = %d W \n ', S_4 , Rot_losses_S_4 );\n",
+"\n",
+"printf(' \n b: Full-load variable loss = %d W\n ',FL_variable_loss );\n",
+"printf(' \n    Variable losses ,');\n",
+"printf(' \n    at %.2f load = %.2f W ',x1 , x1_variable_loss );\n",
+"printf(' \n    at %.2f load = %.2f W ',x2 , x2_variable_loss );\n",
+"printf(' \n    at %.2f load = %.2f W ',x3 , x3_variable_loss );\n",
+"printf(' \n    at %.2f load = %.2f W \n ',x4 , x4_variable_loss );\n",
+"\n",
+"printf(' \n c: Efficiency of motor = (Input - losses)/Input ');\n",
+"printf(' \n    where\n     Input = volts*amperes*load_fraction ');\n",
+"printf(' \n     Losses = field loss + rotational losses + variable electric losses');\n",
+"printf(' \n    Input,\n    at %.2f load = %d W ',x1 , Input_x1 );\n",
+"printf(' \n    at %.2f load = %d W ',x2 , Input_x2 );\n",
+"printf(' \n    at %.2f load = %d W ',x3 , Input_x3 );\n",
+"printf(' \n    at full load = %d W ' , Input_FL );\n",
+"printf(' \n    at %.2f load = %d W \n ',x4  , Input_x4 );\n",
+"\n",
+"printf(' \n    Field loss for each of the conditions of load = %d W \n',Field_loss);\n",
+"printf(' \n    Rotational losses are calculated in part a while variable ');\n",
+"printf(' \n    electric losses in part b \n');\n",
+"\n",
+"printf(' \n    Efficiency at %.2f load = %f = %.1f percent ',x1,eta_1,eta_1*100);\n",
+"printf(' \n    Efficiency at %.2f load = %f = %.1f percent ',x2,eta_2,eta_2*100);\n",
+"printf(' \n    Efficiency at %.2f load = %f = %.1f percent ',x3,eta_3,eta_3*100);\n",
+"printf(' \n    Efficiency at full load = %f = %.1f percent ',eta_FL,eta_FL*100);\n",
+"printf(' \n    Efficiency at %.2f load = %f = %.1f percent \n',x4,eta_4,eta_4*100);\n",
+"\n",
+"printf(' \n d: ________________________________________________________________________________________________________');\n",
+"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   ');\n",
+"printf(' \n    ________________________________________________________________________________________________________');\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);\n",
+"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);\n",
+"printf(' \n\n    Rotational losses');\n",
+"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);\n",
+"printf(' \n\n    Variable electric losses');\n",
+"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);\n",
+"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);\n",
+"printf(' \n    ________________________________________________________________________________________________________');\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);\n",
+"printf(' \n    ________________________________________________________________________________________________________');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.5: Ia_LF_max_efficiency_LF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 10000 ; // Power rating of the shunt generator in W\n",
+"V = 230 ;// Voltage rating of the shunt generator in volt\n",
+"S = 1750 ; // Speed in rpm of the shunt generator\n",
+"R_a = 0.2 ; // Armature resistance\n",
+"// Calculated values from Ex.12-1\n",
+"P_r = 489.2 ; // Shunt generator rotational losses in W\n",
+"Vf_If = 230 ; // Shunt field circuit loss in W\n",
+"I_a_rated = 44.5 ; // Rated armature current in A\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"I_a = sqrt( (Vf_If + P_r) / R_a ); // Armature current in A for max.efficiency\n",
+"\n",
+"// case b\n",
+"LF = I_a / I_a_rated ; // Load fraction\n",
+"LF_percent = LF*100 ; // Load fraction in percent \n",
+"\n",
+"// case c\n",
+"P_k = Vf_If + P_r ;\n",
+"eta_max = (P*LF)/( (P*LF) + (Vf_If + P_r) + P_k ) * 100; // Maximum efficiency\n",
+"\n",
+"// case d\n",
+"// subscript d for LF indicates case d\n",
+"LF_d = sqrt(P_k/(I_a_rated^2*R_a)) ; // Load fraction from fixed losses and rated variable losses\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-5 Solution : ');\n",
+"\n",
+"printf(' \n a: Armature current for max.efficiency :\n    I_a = %.f A \n',I_a);\n",
+"\n",
+"printf(' \n b: Load fraction :\n    L.F. = %.1f percent = %.3f*rated \n',LF_percent,LF);\n",
+"\n",
+"printf(' \n c: Maximum efiiciency :\n    η = %.2f percent \n',eta_max);\n",
+"\n",
+"printf(' \n d: Load fraction from fixed losses and rated variable losses :');\n",
+"printf(' \n    L.F. = %.3f*rated',LF_d);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.6: Pd_Pr_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 240 ; // Voltage rating of dc shunt motor in volt\n",
+"P_hp = 5 ; // Power rating of dc shunt motor in hp\n",
+"S = 1100 ; // Speed in rpm of the dc shunt motor\n",
+"R_a = 0.4 ; // Armture resistance in ohm\n",
+"R_f = 240 ; // Field resistance in ohm\n",
+"I_L = 20 ; // Rated line current in A \n",
+"\n",
+"// Calculations\n",
+"// Preliminary calculations\n",
+"V_f = V ; // Voltage across field winding in volt\n",
+"I_f = V_f / R_f ; // Field current in A\n",
+"I_a = I_L - I_f ; // Armature current in A\n",
+"P_o = P_hp * 746 ; // Power rating of dc shunt motor in W\n",
+"V_a = V ; // Voltage across armature in volt\n",
+"E_c_fl = V_a - I_a*R_a ; // back EMF in volt\n",
+"\n",
+"// case a\n",
+"E_c = E_c_fl ;\n",
+"P_d = E_c * I_a ; // Power developed by the armature in W\n",
+"\n",
+"// case b\n",
+"P_r = P_d - P_o ; // Full-load rotational losses in W\n",
+"\n",
+"// case c\n",
+"P_in = V*I_L ; // Input power in W\n",
+"eta = (P_o/P_in)*100 ; // Full-load efficiency \n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-6 Solution : ');\n",
+"\n",
+"printf(' \n Preliminary calculations using nameplate data : ');\n",
+"printf(' \n Field current : I_f = %d A \n ',I_f);\n",
+"printf(' \n Armature current : I_a = %d A \n ',I_a);\n",
+"printf(' \n P_o = %d W ',P_o );\n",
+"printf(' \n E_c(fl) = %.1f V \n',E_c_fl);\n",
+"\n",
+"printf(' \n a: Power developed by the armature :\n    P_d = %.1f W \n',P_d);\n",
+"\n",
+"printf(' \n b: Full-load rotational losses :\n    P_r = %.1f W \n',P_r);\n",
+"\n",
+"printf(' \n c: Full-laod efficiency :\n    η = %.1f percent ',eta );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.7: Pd_Pr_max_and_fl_efficiency_Pk_Ia_LF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 240 ; // Voltage rating of dc shunt motor in volt\n",
+"P_hp = 25 ; // Power rating of dc shunt motor in hp\n",
+"S = 1100 ; // Speed in rpm of the dc shunt motor\n",
+"R_a = 0.15 ; // Armture resistance in ohm\n",
+"R_f = 80 ; // Field resistance in ohm\n",
+"I_L = 89 ; // Rated line current in A \n",
+"\n",
+"// Calculations\n",
+"// Preliminary calculations\n",
+"V_f = V ; // Voltage across field winding in volt\n",
+"I_f = V_f / R_f ; // Field current in A\n",
+"I_a = I_L - I_f ; // Armature current in A\n",
+"P_o = P_hp * 746 ; // Power rating of dc shunt motor in W\n",
+"V_a = V ; // Voltage across armature in volt\n",
+"E_c_fl = V_a - I_a*R_a ; // back EMF in volt\n",
+"\n",
+"// case a\n",
+"E_c = E_c_fl ;\n",
+"P_d = E_c * I_a ; // Power developed by the armature in W\n",
+"\n",
+"// case b\n",
+"P_r = P_d - P_o ; // Full-load rotational losses in W\n",
+"\n",
+"// case c\n",
+"P_in = V*I_L ; // Input power in W\n",
+"eta_fl = (P_o/P_in)*100 ; // Full-load efficiency \n",
+"\n",
+"// case d\n",
+"P_k = V_f*I_f + P_r ; // Total constant losses in W\n",
+"\n",
+"// case e\n",
+"Ia = sqrt(P_k/R_a); // Armature current in A from maximum efficiency\n",
+"\n",
+"// case f\n",
+"LF = Ia / I_a ; // Load fraction at which max.efficiency is produced\n",
+"\n",
+"// case g\n",
+"rated_input = V*I_L ;\n",
+"eta_max = ( (LF*rated_input) - 2*P_k ) / (LF*rated_input) * 100; // Maximum efficiency\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-7 Solution : ');\n",
+"\n",
+"printf(' \n Field current : I_f = %d A \n ',I_f);\n",
+"printf(' \n Armature current : I_a = %d A \n ',I_a);\n",
+"printf(' \n P_o = %d W \n',P_o );\n",
+"printf(' \n E_c(fl) = %.1f V \n',E_c_fl);\n",
+"\n",
+"printf(' \n a: Power developed by the armature :\n    P_d = %.1f W \n',P_d);\n",
+"\n",
+"printf(' \n b: Full-load rotational losses :\n    P_r = %.1f W \n',P_r);\n",
+"\n",
+"printf(' \n c: Full-laod efficiency :\n    η = %.1f percent \n ',eta_fl );\n",
+"\n",
+"printf(' \n d: Total constant losses :\n    P_k = %.1f W \n',P_k);\n",
+"\n",
+"printf(' \n e: Armature current from maximum efficiency :\n    I_a = %.1f A\n ',Ia);\n",
+"\n",
+"printf(' \n f: L.F. = %.1f \n ',LF);\n",
+"\n",
+"printf(' \n g: η_max = %.1f percent',eta_max);\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.8: IL_Ia_Pd_Pr_Speed_SR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 240 ; // Voltage rating of dc shunt motor in volt\n",
+"P_hp = 5 ; // Power rating of dc shunt motor in hp\n",
+"S_fl = 1100 ; // Speed in rpm of the dc shunt motor\n",
+"R_a = 0.4 ; // Armture resistance in ohm\n",
+"R_f = 240 ; // Field resistance in ohm\n",
+"eta = 0.75 ; // Full-load efficiency\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_L = V ; // Load voltage\n",
+"P_o = P_hp * 746 ; // Power rating of dc shunt motor in W\n",
+"I_L = P_o / (eta*V_L); // Rated input line current in A\n",
+"\n",
+"V_f = V ; // Voltage across field winding in volt\n",
+"I_f = V_f / R_f ; // Field current in A\n",
+"I_a = I_L - I_f ; // Rated armature current in A\n",
+"\n",
+"// case b\n",
+"V_a = V ; // Voltage across armature in volt\n",
+"E_c = V_a - I_a*R_a ; // back EMF in volt\n",
+"P_d = E_c * I_a ; // Power developed by the armature in W\n",
+"\n",
+"// case c\n",
+"P_r = P_d - P_o ; // Rotational losses in W at rated load\n",
+"\n",
+"// case d\n",
+"// At no-load\n",
+"P_o_nl = 0 ;\n",
+"P_r_nl = P_r ; // Rotational losses in W at no load\n",
+"P_d_nl = P_r_nl ;\n",
+"\n",
+"// case e\n",
+"I_a_nl = P_d_nl / V_a ; // No-load armature current in A\n",
+"\n",
+"// case f\n",
+"E_c_nl = V ; // No-load voltage in volt\n",
+"E_c_fl = E_c ; // Full-load voltage in volt\n",
+"S_nl = (E_c_nl / E_c_fl)*S_fl ; // No-load speed in rpm\n",
+"\n",
+"// case g\n",
+"SR = (S_nl - S_fl)/S_fl * 100 ; // Speed regulation\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-8 Solution : ');\n",
+"\n",
+"printf(' \n a: Rated input line current :\n    I_L = %.2f A \n ',I_L);\n",
+"printf(' \n    Rated armature current :\n    I_a = %.2f A \n ',I_a );\n",
+"\n",
+"printf(' \n b: E_c = %.1f V \n ',E_c );\n",
+"printf(' \n    Power developed by the armature at rated load :\n    P_d = %d W \n ',P_d);\n",
+"\n",
+"printf(' \n c: Rotational losses at rated load :\n    P_r = %d W \n ', P_r );\n",
+"\n",
+"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);\n",
+"\n",
+"printf(' \n e: No-load armature current :\n    I_a(nl) = %.2f A \n ',I_a_nl );\n",
+"\n",
+"printf(' \n f: No-load speed :\n    S_nl = %.f rpm \n ',S_nl );\n",
+"\n",
+"printf(' \n g: Speed regulation :\n    SR = %.1f percent ',SR );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 12.9: Ec_Pd_Po_Pr_To_Ia_efficiency_speed_SR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS\n",
+"// Example 12-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 240 ; // Voltage rating of dc shunt motor in volt\n",
+"I_L = 55 ; // Rated line current in A \n",
+"S = 1200 ; // Speed in rpm of the dc shunt motor\n",
+"P_r = 406.4 ; // Rotational losses in W at rated load\n",
+"R_f = 120 ; // Field resistance in ohm\n",
+"R_a = 0.4 ; // Armture resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"\n",
+"V_f = V ; // Voltage across field winding in volt\n",
+"I_f = V_f / R_f ; // Field current in A\n",
+"I_a = I_L - I_f ; // Rated armature current in A\n",
+"\n",
+"V_a = V ; // Voltage across armature in volt\n",
+"E_c = V_a - I_a*R_a ; // back EMF in volt\n",
+"P_d = E_c * I_a ; // Power developed by the armature in W\n",
+"\n",
+"// case b\n",
+"P_o = P_d - P_r  ; // Rated output power in W\n",
+"P_o_hp = P_o / 746 ; // Rated output power in hp\n",
+"\n",
+"// case c\n",
+"T_o = (P_o_hp * 5252)/S ; // C in lb-ft\n",
+"T_o_Nm = T_o * (1.356); // Rated output torque in N-m\n",
+"\n",
+"// case d\n",
+"P_in = V*I_L ; // Input power in W\n",
+"eta = (P_o/P_in)*100 ; // Efficiency at rated load\n",
+"\n",
+"// case e\n",
+"// At no-load\n",
+"P_o_nl = 0 ;\n",
+"P_r_nl = P_r ; // Rotational losses in W at no load\n",
+"P_d_nl = P_r_nl ;\n",
+"\n",
+"I_a_nl = P_d_nl / V_a ; // No-load armature current in A\n",
+"\n",
+"E_c_nl = V ; // No-load voltage in volt\n",
+"E_c_fl = E_c ; // Full-load voltage in volt\n",
+"S_fl = S ; // Full-load speed in rpm\n",
+"S_nl = (E_c_nl / E_c_fl)*S_fl ; // No-load speed in rpm\n",
+"\n",
+"// case f\n",
+"SR = (S_nl - S_fl)/S_fl * 100 ; // Speed regulation\n",
+"\n",
+"// Display the results\n",
+"disp('Example 12-9 Solution : ');\n",
+"\n",
+"printf(' \n a: E_c = %.1f V \n ',E_c );\n",
+"printf(' \n    Power developed by the armature at rated load :\n    P_d = %.1f W \n ',P_d);\n",
+"\n",
+"printf(' \n b: Rated output power :\n    P_o = %d W \n ', P_o );\n",
+"printf(' \n    P_o = %d hp \n ',P_o_hp);\n",
+"\n",
+"printf(' \n c: Rated output torque :\n    T_o = %.2f lb-ft ',T_o);\n",
+"printf(' \n    T_o = %.f N-m \n ',T_o_Nm );\n",
+"\n",
+"printf(' \n d: Efficiency at rated load :\n    η = %.1f percent \n ',eta );\n",
+"\n",
+"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);\n",
+"printf(' \n    No-load armature current :\n    I_a(nl) = %.3f A \n ',I_a_nl );\n",
+"printf(' \n    No-load speed :\n    S_nl = %f ≃ %.f rpm \n ',S_nl,S_nl );\n",
+"\n",
+"printf(' \n f: Speed regulation :\n    SR = %.1f percent ',SR );\n",
+"\n",
+"printf(' \n    Variation in SR is due to non-approximation of S_nl = %f rpm',S_nl);\n",
+"printf(' \n    while calculating SR in scilab .')"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/13-RATINGS_SELECTION_AND_MAINTENANCE_OF_ELECTRIC_MACHINERY.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/13-RATINGS_SELECTION_AND_MAINTENANCE_OF_ELECTRIC_MACHINERY.ipynb
new file mode 100644
index 0000000..49d7f22
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/13-RATINGS_SELECTION_AND_MAINTENANCE_OF_ELECTRIC_MACHINERY.ipynb
@@ -0,0 +1,702 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 13: RATINGS SELECTION AND MAINTENANCE OF ELECTRIC MACHINERY"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.10: new_Zpu.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// single phase alternator\n",
+"V_orig = 500 ; // Rated voltage of the alternator in volt\n",
+"kVA_orig = 20 ; // Rated power of the alternator in kVA\n",
+"I = 40 ; // Rated current of the alternator in A\n",
+"R = 2 ; // Armature resistance in ohm\n",
+"X = 15 ; // Armature reactance in ohm\n",
+"\n",
+"V_new = 5000 ; // New voltage of the alternator in volt\n",
+"kVA_new = 100 ; // New power of the alternator in kVA\n",
+"\n",
+"// Calculated armature impedance from Ex.13-9c\n",
+"Z_pu_orig = 1.211 ; // original per-unit value of armature impedance in p.u\n",
+"\n",
+"// Calculation\n",
+"Z_pu_new = Z_pu_orig * (kVA_new/kVA_orig) * (V_orig/V_new)^2 ;\n",
+"// new per-unit value of armature impedance in p.u\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-10 Solution : ');\n",
+"printf(' \n New per-unit value of armature impedance\n Z_pu(new) = %.5f p.u',Z_pu_new);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.11: line_and_phase_Vpu.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3-phase distribution system\n",
+"V = 2300 ; // Line voltage of 3-phase distribution system in volt\n",
+"V_p = 1328 ; // Phase voltage of 3-phase distribution system in volt\n",
+"\n",
+"V_b = 69000 ; // Common base line voltage in volt\n",
+"V_pb = 39840 ; // Common base phase voltage in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_pu_line = V / V_b ; // Distribution system p.u line voltage\n",
+"\n",
+"// case a\n",
+"V_pu_phase = V_p / V_pb ; // Distribution system p.u phase voltage\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-11 Solution : ');\n",
+"printf(' \n a: Distribution system p.u line voltage :\n    V_pu = %.2f p.u\n',V_pu_line);\n",
+"\n",
+"printf(' \n b: Distribution system p.u phase voltage :\n    V_pu = %.2f p.u\n',V_pu_phase);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.12: Zb_Xs_Ra_Zs_P.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-12\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"VA_b = 50 ; // Base power rating of the 3-phase Y-connected alternator in MVA\n",
+"V_b = 25 ; // Base voltage of the 3-phase Y-connected alternator in kV\n",
+"X_pu = 1.3  ; // per unit value of synchronous reactance\n",
+"R_pu = 0.05  ; // per unit value of resistance\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// subscript 1 for Z_b indicates method 1 for finding Z_b\n",
+"Z_b1 = (V_b)^2 / VA_b ; // Base impedance in ohm\n",
+"\n",
+"// subscript 2 for Z_b indicates method 2 for finding Z_b\n",
+"S_b = VA_b ; // Base power rating of the 3-phase Y-connected alternator in MVA\n",
+"I_b = (S_b)/V_b ; // Base current in kA\n",
+"Z_b2 = V_b / I_b ; // Base impedance in ohm\n",
+"\n",
+"// case b\n",
+"Z_b = Z_b1; // Base impedance in ohm\n",
+"X_s = X_pu * Z_b ; // Actual value of synchronous reactance per phase in ohm\n",
+"\n",
+"// case c\n",
+"R_a = R_pu * Z_b ; // Actual value of armature stator resistance per phase in ohm\n",
+"\n",
+"// case d\n",
+"// subscript 1 for Z_s indicates method 1 for finding Z_s\n",
+"Z_s1 = R_a + %i*X_s ; // Synchronous impedance per phase in ohm\n",
+"Z_s1_m = abs(Z_s1);//Z_s1_m = magnitude of Z_s1 in ohm\n",
+"Z_s1_a = atan(imag(Z_s1) /real(Z_s1))*180/%pi;//Z_s1_a=phase angle of Z_s1 in degrees\n",
+"\n",
+"// subscript 2 for Z_s indicates method 2 for finding Z_s\n",
+"Z_pu = R_pu + %i*X_pu ; // per unit value of impedance\n",
+"Z_s2 = Z_pu * Z_b ; // Synchronous impedance per phase in ohm\n",
+"Z_s2_m = abs(Z_s2);//Z_s2_m = magnitude of Z_s2 in ohm\n",
+"Z_s2_a = atan(imag(Z_s2) /real(Z_s2))*180/%pi;//Z_s2_a=phase angle of Z_s2 in degrees\n",
+"\n",
+"// case e\n",
+"S = S_b ; // Base power rating of the 3-phase Y-connected alternator in MVA\n",
+"P = S * R_pu ; // Full-load copper losses for all three phases in MW\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-12 Solution : ');\n",
+"\n",
+"printf(' \n a: Base impedance(method 1): \n    Z_b = %.1f ohm\n',Z_b1);\n",
+"printf(' \n    Base impedance(method 2) : ');\n",
+"printf(' \n    I_b = %d kA \n    Z_b = %.1f ohm\n',I_b,Z_b2);\n",
+"\n",
+"printf(' \n b: Actual value of synchronous reactance per phase : ');\n",
+"printf(' \n    X_s in ohm = ');disp(%i*X_s);\n",
+"\n",
+"printf(' \n c: Actual value of armature stator resistance per phase : ');\n",
+"printf(' \n    R_a = %.3f ohm \n ',R_a );\n",
+"\n",
+"printf(' \n d: Synchronous impedance per phase (method 1): ');\n",
+"printf(' \n    Z_s in ohm = ');disp(Z_s1);\n",
+"printf(' \n    Z_s = %.2f <%.1f ohm\n',Z_s1_m,Z_s1_a);\n",
+"printf(' \n    Synchronous impedance per phase (method 2) : ');\n",
+"printf(' \n    Z_s in ohm = ');disp(Z_s2);\n",
+"printf(' \n    Z_s = %.2f <%.1f ohm\n',Z_s2_m,Z_s2_a);\n",
+"\n",
+"printf(' \n e: Full-load copper losses for all 3 phases : \n    P = %.1f MW',P);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.1: R_and_reduced_life_expectancy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// MOTOR(class A insulation ) is operated for 6 hrs\n",
+"T = 125 ; // Temperature in degree celsius recorded by the embedded detectors\n",
+"life_orig = 10 ; // Life in years of the motor (standard)\n",
+"\n",
+"// Calculations\n",
+"delta_T = T - 105 ; // Positive temperature difference between the given\n",
+"// max hottest spot temperature of its insulation and the ambient temperature recorded.\n",
+"// 105 is chosen from table 13-1(class A insulation)\n",
+"R = 2 ^ (delta_T/10); // Life reduction factor\n",
+"  \n",
+"Life_calc = life_orig / R ; // Reduced life expectancy of the motor in years\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-1 Solution : ');\n",
+"printf(' \n Life reduction factor : R = %d \n  ',R );\n",
+"printf(' \n Reduced life expectancy of the motor : Life_calc = %.1f years',Life_calc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.2: E_and_increased_life_expectancy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// MOTOR(class A insulation ) is operated for 6 hrs\n",
+"T = 75 ; // Temperature in degree celsius recorded by the embedded detectors\n",
+"life_orig = 10 ; // Life in years of the motor (standard)\n",
+"\n",
+"// Calculations\n",
+"delta_T = 105 - T ; // Positive temperature difference between the given\n",
+"// max hottest spot temperature of its insulation and the ambient temperature recorded.\n",
+"// 105 is chosen from table 13-1 (class A insulation)\n",
+"E = 2 ^ (delta_T/10); // Life extension factor\n",
+"  \n",
+"Life_calc = life_orig * E ; // Increased life expectancy of the motor in years\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-2 Solution : ');\n",
+"printf(' \n Life extension factor : E = %d \n ',E );\n",
+"printf(' \n Increased life expectancy of the motor : Life_calc = %d years ',Life_calc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.3: E_and_increased_life_expectancy_classB.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Class A insulation\n",
+"T_A = 105 ; // Temperature in degree celsius recorded by the embedded detectors\n",
+"life_orig = 5 ; // Life in years of the motor (standard)\n",
+"// Class B insulation\n",
+"T_B = 130 ; // Temperature in degree celsius recorded by the embedded detectors\n",
+"\n",
+"// Calculations\n",
+"delta_T = T_B - T_A ; // Positive temperature difference betw the given\n",
+"// max hottest spot temperature of its insulation and the ambient temperature recorded.\n",
+"// T_A and T_B are chosen from table 13-1\n",
+"E = 2 ^ (delta_T/10); // Life extension factor\n",
+"  \n",
+"Life_calc = life_orig * E ; // Increased life expectancy of the motor in years\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-3 Solution : ');\n",
+"printf(' \n Life extension factor : E = %.2f \n ',E );\n",
+"printf(' \n Increased life expectancy of the motor : Life_calc = %.1f years ',Life_calc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.4: ClassB_insulation_SCIM_details.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 25 ; // Rated power of SCIM in hp\n",
+"// class B insulation\n",
+"T_ambient = 40 ; // Standard ambient temperature recorded by the embedded hot-spot detectors in degree celsius\n",
+"T_hottest = 115 ; // Hottest-spot winding temperature recorded by the embedded hot-spot detectors in degree celsius\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// from table 13-1 allowable temperature rise in 90 degree celsius\n",
+"\n",
+"// case b\n",
+"T_rise = T_hottest - T_ambient ; // Actual temperature rise for the insulation type used in degree celsius\n",
+"\n",
+"// case c\n",
+"P_f = P_o * (90/T_rise); // Approximate power to the motor that can be delivered at T_rise\n",
+"\n",
+"// case d\n",
+"// same as P_f\n",
+"\n",
+"// case e\n",
+"// answer from case a\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-4 Solution : ');\n",
+"printf(' \n a: The allowable temperature rise for the ');\n",
+"printf(' \n    insulation type used = 90 degree celsius(from table 13-1)\n');\n",
+"\n",
+"printf(' \n b: The actual temperature rise for the insulation type used = %d degree celsius\n',T_rise);\n",
+"\n",
+"printf(' \n c: The approximate power to the motor that can be delivered at T_rise');\n",
+"printf(' \n    P_f = %d hp\n',P_f);\n",
+"\n",
+"printf(' \n d: Power rating that may be stamped on the nameplate = %d hp(subject to test at this load) \n ',P_f);\n",
+"\n",
+"printf(' \n e: The temperature rise that must be stamped on the nameplate = 90 degree celsius');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.5: final_temperature.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 50 ; // Power rating of the WRIM in hp\n",
+"// Class F insulation\n",
+"T_hottest = 160 ; // Hottest-spot winding temperature recorded by the embedded \n",
+"// hot-spot detectors in degree celsius\n",
+"T_ambient = 40 ; // Standard ambient temperature recorded by the embedded \n",
+"// hot-spot detectors in degree celsius\n",
+"P_f_a = 40 ; // Power rating of load a in hp\n",
+"P_f_b = 55 ; // Power rating of load a in hp\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"delta_T_o = T_hottest - T_ambient ; // Temperature rise for the insulation type \n",
+"// used in degree celsius\n",
+"\n",
+"// subscript a in delta_T_f ,P_f_a and T_f indicates case a\n",
+"delta_T_f_a = (P_f_a/P_o)*delta_T_o ; // final temperature rise in degree celsius\n",
+"T_f_a = delta_T_f_a + T_ambient ; // Approximate final hot-spot temperature in degree celsius\n",
+"\n",
+"// case b\n",
+"// subscript b in delta_T_f ,P_f and T_f indicates case b\n",
+"delta_T_f_b = (P_f_b/P_o)*delta_T_o ; // final temperature rise in degree celsius\n",
+"T_f_b = delta_T_f_b + T_ambient ; // Approximate final hot-spot temperature in degree celsius\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-5 Solution : ');\n",
+"printf(' \n a: ΔT_o = %d degree celsius ',delta_T_o);\n",
+"printf(' \n    ΔT_f = %d degree celsius ',delta_T_f_a);\n",
+"printf(' \n    T_f = %d degree celsius \n',T_f_a);\n",
+"\n",
+"printf(' \n b: ΔT_f = %d degree celsius ',delta_T_f_b);\n",
+"printf(' \n    T_f = %d degree celsius \n',T_f_b);\n",
+"printf(' \n    Yes,motor life is reduced at the 110 percent motor load because');\n",
+"printf(' \n    the allowable maximum hot-spot motor temperature for Class F');\n",
+"printf(' \n    insulation is 155 degree celsius.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.6: Tf_R_decreased_life_expectancy.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 55 ; // Power rating of the WRIM in hp\n",
+"T_ambient = 40 ; // Standard ambient temperature recorded by the embedded \n",
+"// hot-spot detectors in degree celsius\n",
+"life_orig = 10 ; // Life in years of the motor (standard)\n",
+"\n",
+"// Calculated data from Ex.13-5b\n",
+"T_f = 172 ; // Approximate final hot-spot temperature in degree celsius\n",
+"\n",
+"// Calculations\n",
+"delta_T = T_f - 155 ; // Positive temperature difference betw the given\n",
+"// max hottest spot temperature of its insulation and the ambient temperature recorded.\n",
+"// 155 is chosen from table 13-1(class F insulation)\n",
+"\n",
+"R = 2 ^ (delta_T/10); // Life reduction factor\n",
+"  \n",
+"Life_calc = life_orig / R ; // Reduced life expectancy of the motor in years\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-6 Solution : ');\n",
+"printf(' \n From Ex.13-5b,T_f = %d degree celsius\n',T_f);\n",
+"printf(' \n Life reduction factor : R = %.2f \n  ',R );\n",
+"printf(' \n Reduced life expectancy of the motor : Life_calc = %.2f years',Life_calc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.7: rms_hp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 200 ; // Power rating of the test motor in hp\n",
+"t1 = 5 ; // time duration in minutes for which test motor is operated at 200 hp \n",
+"t2 = 5 ; // time duration in minutes for which test motor is operated at 20 hp \n",
+"t3 = 10 ; // time duration in minutes for which test motor is operated at 100 hp \n",
+"\n",
+"// Calculation\n",
+"rms_hp = sqrt( ( (200^2)*t1 + (20^2)*t2 + (100^2)*t3 )/(t1 + t2 + t3 + 10/3) );\n",
+"// Horsepower required for intermittent varying load\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-7 Solution : ');\n",
+"printf(' \n Horsepower required for intermittent varying load is : ');\n",
+"printf(' \n rms hp = %.f hp \n ',rms_hp);\n",
+"\n",
+"printf(' \n A 125 hp motor would be selected because that is the nearest larger');\n",
+"printf(' \n commercial standard rating.This means that the motor would operate ');\n",
+"printf(' \n with a 160 percent overload (at 200 hp) for 5 minutes,or 1/6th of ')\n",
+"printf(' \n its total duty cycle.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.8: Vb_Ib_Rb_Rpu.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 120 ; // Rated output voltage in volt of separately excited dc generator\n",
+"I = 100 ; // Rated output current in A of separately excited dc generator\n",
+"R = 0.1 ; // Armature resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_b = V ; // base voltage in volt\n",
+"\n",
+"// case b\n",
+"I_b = I ; // base current in A\n",
+"\n",
+"// case c\n",
+"R_b = V_b / I_b ; // base resistance in ohm\n",
+"\n",
+"// case d\n",
+"R_pu = R / R_b ; // per-unit value of armature resistance in p.u\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-8 Solution : ');\n",
+"\n",
+"printf(' \n a: Base voltage \n    V_b = %d V \n ', V_b );\n",
+"\n",
+"printf(' \n b: Base current \n    I_b = %d A \n ', I_b );\n",
+"\n",
+"printf(' \n c: Base resistance \n    R_b = %.1f ohm \n ', R_b );\n",
+"\n",
+"printf(' \n d: Per-unit value of armature resistance\n    R_p.u = %.3f p.u \n ', R_pu );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 13.9: Rpu_jXpu_Zpu.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY\n",
+"// Example 13-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// single phase alternator\n",
+"V = 500 ; // Rated voltage of the alternator in volt\n",
+"P = 20 ; // Rated power of the alternator in kVA\n",
+"I = 40 ; // Rated current of the alternator in A\n",
+"R = 2 ; // Armature resistance in ohm\n",
+"X = 15 ; // Armature reactance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_b = V ; // base voltage in volt\n",
+"I_b = I ; // base current in A\n",
+"R_pu = (R*I_b)/V_b ; // per-unit value of armature resistance in p.u\n",
+"\n",
+"// case b\n",
+"jX_pu = (X*I_b)/V_b ; // per-unit value of armature reactance in p.u\n",
+"\n",
+"// case c\n",
+"// subscript 1 indicates method 1 for finding Z_p.u\n",
+"Z_pu1 = R_pu + %i*(jX_pu); // per-unit value of armature impedance in p.u\n",
+"Z_pu1_m = abs(Z_pu1);//Z_pu1_m = magnitude of Z_pu1 in p.u\n",
+"Z_pu1_a = atan(imag(Z_pu1) /real(Z_pu1))*180/%pi;//Z_pu1_a=phase angle of Z_pu1 in degrees\n",
+"\n",
+"// subscript 2 indicates method 2 for finding Z_p.u\n",
+"Z_pu2 = (R + %i*X)*(I/V); // per-unit value of armature impedance in p.u\n",
+"Z_pu2_m = abs(Z_pu2);//Z_pu2_m = magnitude of Z_pu2 in p.u\n",
+"Z_pu2_a = atan(imag(Z_pu2) /real(Z_pu2))*180/%pi;//Z_pu2_a=phase angle of Z_pu2 in degrees\n",
+"\n",
+"// Display the results\n",
+"disp('Example 13-9 Solution : ');\n",
+"\n",
+"printf(' \n a: Armature resistance per unit value\n    R_p.u = %.2f p.u \n',R_pu);\n",
+"\n",
+"printf(' \n b: Armature reactance per unit value\n    jX_p.u in p.u = ');disp(%i*jX_pu);\n",
+"\n",
+"printf(' \n c: Armature impedance per unit value\n');\n",
+"printf(' \n    (method 1)\n    Z_p.u in p.u = ');disp(Z_pu1);\n",
+"printf(' \n    Z_p.u = %.3f <%.1f p.u \n',Z_pu1_m,Z_pu1_a );\n",
+"\n",
+"printf(' \n    (method 2)\n    Z_p.u in p.u = ');disp(Z_pu2);\n",
+"printf(' \n    Z_p.u = %.3f <%.1f p.u \n',Z_pu2_m,Z_pu2_a );"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/14-TRANSFORMERS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/14-TRANSFORMERS.ipynb
new file mode 100644
index 0000000..a6935ac
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/14-TRANSFORMERS.ipynb
@@ -0,0 +1,3305 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 14: TRANSFORMERS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.10: Piro.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data(from Example 14-9)\n",
+"kVA = 1 ; // kVA rating of the transformer\n",
+"V_1 = 220 ; // Primary voltage in volt\n",
+"V_2 = 110 ; // Secondary voltage in volt\n",
+"f_o = 400 ; // Frequency in Hz\n",
+"f_f = 60 ; // Frequency in Hz for which the transformer is to be used\n",
+"P_orig = 10 ; // Original iron losses of the transformer in W\n",
+"\n",
+"// Calculations\n",
+"// consider only ratio of frequencies for calculating B  \n",
+"B = f_o / f_f ; // flux density\n",
+"\n",
+"P_iron = (P_orig)*(B^2); // Iron losses in W\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-10 Solution : ');\n",
+"\n",
+"printf(' \n Since E = k*f*B_m and the same primary voltage is applied to the ');\n",
+"printf(' \n transformer at reduced frequency, the final flux density B_mf ');\n",
+"printf(' \n increased significantly above its original maximum permissible ');\n",
+"printf(' \n value B_mo to :\n B_mf = B_mo * (f_o/f_f) = %.2fB_mo \n ',B );\n",
+"\n",
+"printf(' \n Since the iron losses vary approximately as the square of the flux density :');\n",
+"printf(' \n P_iron = %d W ',P_iron );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.11: I2_I1_Z2_Z1their_loss_E2_E1_alpha.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 500 ; // kVA rating of the step-down transformer\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"r_1 = 0.1 ; // Primary winding resistance in ohm\n",
+"x_1 = 0.3 ; // Primary winding reactance in ohm\n",
+"r_2 = 0.001 ; // Secondary winding resistance in ohm\n",
+"x_2 = 0.003 ; // Secondary winding reactance in ohm\n",
+"\n",
+"// Calculations\n",
+"alpha = V_1 / V_2 ; // Transformation ratio\n",
+"// case a\n",
+"I_2 = (kVA*1000) / V_2 ; // Secondary current in A\n",
+"I_1 = I_2 / alpha ; // Primary current in A\n",
+"\n",
+"// case b\n",
+"Z_2 = r_2 + %i*(x_2); // Secondary internal impedance in ohm\n",
+"Z_2_m = abs(Z_2);//Z_2_m=magnitude of Z_2 in ohm\n",
+"Z_2_a = atan(imag(Z_2) /real(Z_2))*180/%pi;//Z_2_a=phase angle of Z_2 in degrees\n",
+"\n",
+"Z_1 = r_1 + %i*(x_1); // Primary internal impedance in ohm\n",
+"Z_1_m = abs(Z_1);//Z_1_m=magnitude of Z_1 in ohm\n",
+"Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees\n",
+"\n",
+"// case c\n",
+"I_2_Z_2 = I_2 * Z_2_m ; // Secondary internal voltage drop in volt\n",
+"I_1_Z_1 = I_1 * Z_1_m ; // Primary internal voltage drop in volt\n",
+"\n",
+"// case d\n",
+"E_2 = V_2 + I_2_Z_2 ; // Secondary induced voltage in volt\n",
+"E_1 = V_1 - I_1_Z_1 ; // Primary induced voltage in volt\n",
+"\n",
+"// case e\n",
+"ratio_E = E_1 / E_2 ; // ratio of primary to secondary induced voltage\n",
+"ratio_V = V_1 / V_2 ; // ratio of primary to secondary terminal voltage\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-11 Solution : ');\n",
+"\n",
+"printf(' \n a: Secondary current :\n    I_2 = %.f A \n ',I_2 );\n",
+"printf(' \n    Primary current :\n    I_1 = %.1f A \n ',I_1 );\n",
+"\n",
+"printf(' \n b: Secondary internal impedance : \n    Z_2 in ohm = ');disp(Z_2);\n",
+"printf(' \n    Z_2 = %f <%.2f ohm \n ',Z_2_m , Z_2_a );\n",
+"printf(' \n    Primary internal impedance : \n    Z_1 in ohm = ');disp(Z_1);\n",
+"printf(' \n    Z_1 = %f <%.2f ohm \n ',Z_1_m , Z_1_a );\n",
+"\n",
+"printf(' \n c: Secondary internal voltage drop :\n    I_2*Z_2 = %.2f V \n ',I_2_Z_2);\n",
+"printf(' \n    Primary internal voltage drop :\n    I_1*Z_1 = %.2f V \n ',I_1_Z_1);\n",
+"\n",
+"printf(' \n d: Secondary induced voltage :\n    E_2 = %.2f V \n',E_2 );\n",
+"printf(' \n    Primary induced voltage :\n    E_1 = %.2f V \n',E_1 );\n",
+"\n",
+"printf(' \n e: Ratio of E_1/E_2 = %.2f = α = N_1/N_2 \n',ratio_E );\n",
+"printf(' \n    But V_1/V_2 = %d ',ratio_V );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.12: ZL_ZP_difference.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-12\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data(from Example 14-11)\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"I_2 = 2174 ; // Secondary current in A\n",
+"I_1 = 217.4 ; // Primary current in A\n",
+"// calculated values from Example 14-11\n",
+"Z_2 = 0.00316 ; // Secondary internal impedance in ohm\n",
+"Z_1 = 0.316 ; // Primary internal impedance in ohm\n",
+"\n",
+"\n",
+"// Calculations\n",
+"alpha = V_1 / V_2 ; // Transformation ratio\n",
+"// case a\n",
+"Z_L = V_2 / I_2 ; // Load impedance in ohm\n",
+"\n",
+"// case b\n",
+"Z_p = V_1 / I_1 ; // Primary input impedance in ohm\n",
+"\n",
+"Zp = (alpha)^2 * Z_L ; // Primary input impedance in ohm\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-12 Solution : ');\n",
+"\n",
+"printf(' \n a: Load impedance :\n    Z_L = %.4f ohm \n ', Z_L );\n",
+"\n",
+"printf(' \n b: Primary input impedance : ');\n",
+"printf(' \n    (method 1) :\n    Z_p = %.2f ohm \n ',Z_p );\n",
+"printf(' \n    (method 2) :\n    Z_p = %.2f ohm \n ',Zp );\n",
+"\n",
+"printf(' \n c: The impedance of the load Z_L = %.4f Ω, which is much greater',Z_L);\n",
+"printf(' \n    than the internal secondary impedance Z_2 = %.5f Ω .\n ',Z_2);\n",
+"printf(' \n    The primary input impedance Z_p = %.2f Ω,which is much greater',Z_p);\n",
+"printf(' \n    than the internal primary impedance Z_1 = %.3f Ω .\n',Z_1);\n",
+"\n",
+"printf(' \n d: It is essential for Z_L to be much greater than Z_2 so that the ');\n",
+"printf(' \n    major part of the voltage produced by E_2 is dropped across the ');\n",
+"printf(' \n    load impedance Z_L. As Z_L is reduced in proportion to Z_2, the ');\n",
+"printf(' \n    load current increases and more voltage is dropped internally ');\n",
+"printf(' \n    across Z_2.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.13: Re1_Xe1_Ze1_ZLprime_I1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-13\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 500 ; // kVA rating of the step-down transformer\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"r_1 = 0.1 ; // Primary winding resistance in ohm\n",
+"x_1 = 0.3 ; // Primary winding reactance in ohm\n",
+"r_2 = 0.001 ; // Secondary winding resistance in ohm\n",
+"x_2 = 0.003 ; // Secondary winding reactance in ohm\n",
+"// calculated data from Example 14-12\n",
+"Z_L = 0.1058 ; // Load impedance in ohm\n",
+"\n",
+"// Calculations\n",
+"alpha = V_1 / V_2 ; // Transformation ratio\n",
+"\n",
+"// case a\n",
+"R_e1 = r_1 + (alpha)^2 * r_2 ; // Equivalent internal resistance referred to the\n",
+"// primary side in ohm\n",
+"\n",
+"// case b\n",
+"X_e1 = x_1 + (alpha)^2 * x_2 ; // Equivalent internal reactance referred to the\n",
+"// primary side in ohm\n",
+"\n",
+"// case c\n",
+"Z_e1 = R_e1 + %i*(X_e1) ; // Equivalent internal impedance referred to the\n",
+"// primary side in ohm\n",
+"Z_e1_m = abs(Z_e1);//Z_e1_m=magnitude of Z_e1 in ohm\n",
+"Z_e1_a = atan(imag(Z_e1) /real(Z_e1))*180/%pi;//Z_e1_a=phase angle of Z_e1 in degrees\n",
+"\n",
+"// case d\n",
+"Z_L_prime = (alpha)^2 * (Z_L); // Equivalent secondary load impedance referred\n",
+"//  to the primary side in ohm\n",
+"\n",
+"// case e\n",
+"R_L = Z_L ; // Load resistance in ohm\n",
+"X_L = 0 ; // Load reactance in ohm\n",
+"\n",
+"// Primary load current in A , when V_1 = 2300 V\n",
+"I_1 = V_1 / ( (R_e1 + alpha^2*R_L) + %i*(X_e1 + alpha^2*X_L) ); \n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-13 Solution : ');\n",
+"\n",
+"printf(' \n a: Equivalent internal resistance referred to the primary side :');\n",
+"printf(' \n    R_c1 = %.2f ohm \n ',R_e1 );\n",
+"\n",
+"printf(' \n b: Equivalent internal reactance referred to the primary side :');\n",
+"printf(' \n    X_c1 = %.2f ohm \n ',X_e1 );\n",
+"\n",
+"printf(' \n c: Equivalent internal impedance referred to the primary side : ');\n",
+"printf(' \n    Z_c1 in ohm = ');disp(Z_e1);\n",
+"printf(' \n    Z_c1 = %.3f <%.2f ohm \n ', Z_e1_m , Z_e1_a );\n",
+"\n",
+"printf(' \n d: Equivalent secondary load impedance referred to the primary side :');\n",
+"printf(' \n    (alpha)^2 * Z_L = %.2f ohm = (alpha)^2 * R_L \n',Z_L_prime);\n",
+"\n",
+"printf(' \n e: Primary load current :\n    I_1 = %f A ≈ %.f A ', I_1, I_1);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.14: I2_ohmdrops_E2_VR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-14\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 500 ; // kVA rating of the step-down transformer\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"R_e2 = 2 ; // Equivalent resistance referred to the\n",
+"// primary side in mΩ\n",
+"X_e2 = 6 ; // Equivalent reactance referred to the\n",
+"// primary side in mΩ\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"I_2 = (kVA ) / V_2 ; // Rated secondary current in kA\n",
+"\n",
+"// case b\n",
+"R_e2_drop = I_2 * R_e2 ; // Full-load equivalent resistance voltage drop in volt\n",
+"\n",
+"// case c\n",
+"X_e2_drop = I_2 * X_e2 ; // Full-load equivalent reactance voltage drop in volt\n",
+"\n",
+"// case d\n",
+"// unity PF \n",
+"cos_theta2 = 1;\n",
+"sin_theta2 = sqrt(1 - (cos_theta2)^2);\n",
+"\n",
+"// Induced voltage when the transformer is delivering rated current to unity PF load\n",
+"E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 + I_2*X_e2);\n",
+"E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt\n",
+"E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees\n",
+"\n",
+"// case e\n",
+"VR = ( (E_2_m - V_2) / V_2 ) * 100 ; // Percent voltage regulation at unity PF \n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-14 Solution : ');\n",
+"\n",
+"printf(' \n a: Rated secondary current :\n    I_2 = %.3f kA \n ', I_2 );\n",
+"\n",
+"printf(' \n b: Full-load equivalent resistance voltage drop : ');\n",
+"printf(' \n    I_2*R_c2 = %.2f V \n', R_e2_drop );\n",
+"\n",
+"printf(' \n c: Full-load equivalent reactance voltage drop : ');\n",
+"printf(' \n    I_2*X_c2 = %.2f V \n', X_e2_drop );\n",
+"\n",
+"printf(' \n d: Induced voltage when the transformer is delivering rated current ');\n",
+"printf(' \n    to unity PF load :\n    E_2 in volt = ');disp(E_2);\n",
+"printf(' \n    E_2 = %.2f <%.2f V \n ',E_2_m , E_2_a);\n",
+"\n",
+"printf(' \n e: Voltage regulation at unity PF :\n    VR = %.2f percent ',VR );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.15: E2_VR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-15\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 500 ; // kVA rating of the step-down transformer\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"R_e2 = 2 ; // Equivalent resistance referred to the\n",
+"// primary side in mΩ\n",
+"X_e2 = 6 ; // Equivalent reactance referred to the\n",
+"// primary side in mΩ\n",
+"I_2 = 2.174 ; //  Rated secondary current in kA\n",
+"\n",
+"cos_theta2 = 0.8 ; // lagging PF\n",
+"sin_theta2 = sqrt(1 - (cos_theta2)^2);\n",
+"\n",
+"// Calculations\n",
+"\n",
+"// case d\n",
+"// Induced voltage when the transformer is delivering rated current to 0.8 lagging PF load\n",
+"E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 + I_2*X_e2);\n",
+"E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt\n",
+"E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees\n",
+"\n",
+"// case e\n",
+"VR = ( (E_2_m - V_2) / V_2 ) * 100 ; // Percent voltage regulation at 0.8 PF lag\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-15 Solution : ');\n",
+"\n",
+"printf(' \n d: Induced voltage when the transformer is delivering rated current ');\n",
+"printf(' \n    to 0.8 lagging PF load :\n    E_2 in volt = ');disp(E_2);\n",
+"printf(' \n    E_2 = %.2f <%.2f V \n ',E_2_m , E_2_a);\n",
+"\n",
+"printf(' \n e: Voltage regulation at 0.8 lagging PF :\n    VR = %.2f percent ',VR );\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.16: E2_VR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-16\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 500 ; // kVA rating of the step-down transformer\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"R_e2 = 2 ; // Equivalent resistance referred to the\n",
+"// primary side in mΩ\n",
+"X_e2 = 6 ; // Equivalent reactance referred to the\n",
+"// primary side in mΩ\n",
+"I_2 = 2.174 ; //  Rated secondary current in kA\n",
+"\n",
+"cos_theta2 = 0.6 ; // leading PF\n",
+"sin_theta2 = sqrt(1 - (cos_theta2)^2);\n",
+"\n",
+"// Calculations\n",
+"\n",
+"// case d\n",
+"// Induced voltage when the transformer is delivering rated current to unity PF load\n",
+"E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 - I_2*X_e2);\n",
+"E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt\n",
+"E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees\n",
+"\n",
+"// case e\n",
+"VR = ( (E_2_m - V_2) / V_2 ) * 100 ; // Percent voltage regulation at 0.8 leading PF \n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-16 Solution : ');\n",
+"\n",
+"printf(' \n d: Induced voltage when the transformer is delivering rated current ');\n",
+"printf(' \n    to 0.6 leading PF load :\n    E_2 in volt = ');disp(E_2);\n",
+"printf(' \n    E_2 = %.2f <%.2f V \n ',E_2_m , E_2_a);\n",
+"\n",
+"printf(' \n e: Voltage regulation at 0.8 leading PF :\n    VR = %.2f percent ',VR );\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.17: Ze1_Re1_Xe1_Ze2_Re2_Xe2their_drops_VR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-17\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 20 ; // kVA rating of the step-down transformer\n",
+"S = 20000 ; // power rating of the step-down transformer in VA\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"\n",
+"// w.r.t HV side following is SC-test data\n",
+"P1 = 250 ; // wattmeter reading in W\n",
+"I1 = 8.7 ; // Input current in A\n",
+"V1 = 50 ; // Input voltage in V\n",
+"\n",
+"// Calculations\n",
+"alpha = V_1 / V_2 ; // Transformation ratio\n",
+"// case a\n",
+"Z_e1 = V1 / I1 ; // Equivalent impedance w.r.t HV side in ohm\n",
+"\n",
+"R_e1 = P1 / (I1)^2 ; // Equivalent resistance w.r.t HV side in ohm\n",
+"\n",
+"theta = acosd(R_e1/Z_e1) ; // PF angle in degrees\n",
+"\n",
+"X_e1 = Z_e1*sind(theta); // Equivalent reactance w.r.t HV side in ohm\n",
+"\n",
+"// case b\n",
+"Z_e2 = Z_e1 / (alpha)^2 ; // Equivalent impedance w.r.t LV side in ohm\n",
+"\n",
+"R_e2 = R_e1 / (alpha)^2 ; // Equivalent resistance w.r.t LV side in ohm\n",
+"\n",
+"X_e2 = Z_e2*sind(theta); // Equivalent reactance w.r.t LV side in ohm\n",
+"\n",
+"// case c\n",
+"I_2 = S / V_2 ; // Rated secondary load current in A\n",
+"\n",
+"R_e2_drop = I_2 * R_e2 ; // Full-load equivalent resistance voltage drop in volt\n",
+"X_e2_drop = I_2 * X_e2 ; // Full-load equivalent reactance voltage drop in volt\n",
+"\n",
+"// At unity PF\n",
+"cos_theta2 = 1;\n",
+"sin_theta2 = sqrt(1 - (cos_theta2)^2);\n",
+"\n",
+"// Induced voltage when the transformer is delivering rated current to unity PF load\n",
+"E_2 = (V_2*cos_theta2 + I_2*R_e2) + %i*(V_2*sin_theta2 + I_2*X_e2);\n",
+"E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt\n",
+"E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees\n",
+"\n",
+"VR_unity_PF = ( (E_2_m - V_2) / V_2 ) * 100 ; // Transformer voltage regulation\n",
+"\n",
+"// case d\n",
+"// at 0.7 lagging PF\n",
+"cos_theta_2 = 0.7 ; // lagging PF\n",
+"sin_theta_2 = sqrt(1 - (cos_theta_2)^2);\n",
+"\n",
+"// Induced voltage when the transformer is delivering rated current to unity PF load\n",
+"E2 = (V_2*cos_theta_2 + I_2*R_e2) + %i*(V_2*sin_theta_2 + I_2*X_e2);\n",
+"E2_m = abs(E2);//E2_m=magnitude of E2 in volt\n",
+"E2_a = atan(imag(E2) /real(E2))*180/%pi;//E2_a=phase angle of E2 in degrees\n",
+"\n",
+"VR_lag_PF = ( (E2_m - V_2) / V_2 ) * 100 ; // Transformer voltage regulation\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-17 Solution : ');\n",
+"\n",
+"printf(' \n a: Equivalent impedance w.r.t HV side :\n    Z_e1 = %.2f Ω \n',Z_e1);\n",
+"printf(' \n    Equivalent resistance w.r.t HV side :\n    R_e1 = %.1f Ω \n',R_e1);\n",
+"printf(' \n    θ = %.f degrees \n',theta );\n",
+"printf(' \n    Equivalent reactance w.r.t HV side :\n    X_e1 = %.2f \n',X_e1);\n",
+"\n",
+"printf(' \n b: Equivalent impedance w.r.t LV side :');\n",
+"printf(' \n    Z_e2 = %f Ω = %.2f mΩ \n',Z_e2 ,Z_e2*1000);\n",
+"printf(' \n    Equivalent resistance w.r.t LV side :\n    R_e2 = %f Ω \n',R_e2);\n",
+"printf(' \n    R_e2 = %f Ω = %.2f mΩ \n',R_e2,R_e2*1000);\n",
+"printf(' \n    Equivalent reactance w.r.t LV side :\n    X_e2 = %f Ω \n',X_e2);\n",
+"printf(' \n    X_e2 = %f Ω = %.2f mΩ \n',X_e2,X_e2*1000);\n",
+"\n",
+"printf(' \n c: Rated secondary load current :\n    I_2 = %.f A\n',I_2);\n",
+"printf(' \n    I_2*R_c2 = %.2f V \n', R_e2_drop );\n",
+"printf(' \n    I_2*X_c2 = %.2f V \n', X_e2_drop );\n",
+"printf(' \n    At unity PF,\n    E_2 in volt = ');disp(E_2);\n",
+"printf(' \n    E_2 = %.2f <%.2f V \n ',E_2_m , E_2_a);\n",
+"printf(' \n    Voltage regulation at unity PF :\n    VR = %.2f percent ',VR_unity_PF );\n",
+"\n",
+"printf(' \n\n d: At 0.7 lagging PF, \n    E_2 in volt = ');disp(E2);\n",
+"printf(' \n    E_2 = %.2f <%.2f V \n ',E2_m , E2_a);\n",
+"printf(' \n    Voltage regulation at 0.7 lagging PF :\n    VR = %.2f percent ',VR_lag_PF );\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.18: Pcsc.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-18\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_sc = 50 ; // Short circuit voltage in volt\n",
+"V_1 = 2300 ; // Rated primary voltage in volt\n",
+"\n",
+"// Calculations\n",
+"P_c = poly(0,'P_c');// Making P_c as a variable just for displaying answer as per\n",
+"// textbook\n",
+"\n",
+"P_c_sc = (V_sc / V_1)^2 * P_c ; // Fraction of P_c measured by the wattmeter\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-18 Solution : ');\n",
+"\n",
+"printf(' \n Since P_c is proportional to the square of the primary voltage V_sc, ');\n",
+"printf(' \n then under short circuit conditions,the fraction of rated-core loss is :');\n",
+"printf(' \n P_c(sc) = ');disp(P_c_sc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.19: Ze1drop_Re1drop_Xe1drop_VR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-19\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"\n",
+"kVA = 20 ; // kVA rating of the step-down transformer\n",
+"S = 20000 ; // power rating of the step-down transformer in VA\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"Z_e1 = 5.75 ; // Equivalent impedance w.r.t HV side in ohm\n",
+"R_e1 = 3.3 ; // Equivalent resistance w.r.t HV side in ohm\n",
+"X_e1 = 4.71 ; // Equivalent reactance w.r.t HV side in ohm\n",
+"\n",
+"// w.r.t HV side following is SC-test data\n",
+"P1 = 250 ; // wattmeter reading in W\n",
+"I1 = 8.7 ; // Input current in A\n",
+"V1 = 50 ; // Input voltage in V\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"Z_e1_drop = V1 ; // High voltage impedance drop in volt\n",
+"\n",
+"// case b\n",
+"theta = acosd(R_e1/Z_e1) ; // PF angle in degrees\n",
+"\n",
+"R_e1_drop = I1*Z_e1*cosd(theta) ; //HV-side equivalent resistance voltage drop in volt\n",
+"\n",
+"// case c\n",
+"X_e1_drop = I1*Z_e1*sind(theta) ; //HV-side equivalent reactance voltage drop in volt\n",
+"\n",
+"// case d\n",
+"// At unity PF\n",
+"cos_theta1 = 1;\n",
+"sin_theta1 = sqrt(1 - (cos_theta1)^2);\n",
+"\n",
+"// Induced voltage when the transformer is delivering rated current to unity PF load\n",
+"E_1 = (V_1*cos_theta1 + I1*R_e1) + %i*(V_1*sin_theta1 + I1*X_e1);\n",
+"E_1_m = abs(E_1);//E_1_m=magnitude of E_1 in volt\n",
+"E_1_a = atan(imag(E_1) /real(E_1))*180/%pi;//E_1_a=phase angle of E_1 in degrees\n",
+"\n",
+"VR_unity_PF = ( (E_1_m - V_1) / V_1 ) * 100 ; // Transformer voltage regulation\n",
+"\n",
+"// case e\n",
+"// at 0.7 lagging PF\n",
+"cos_theta_1 = 0.7 ; // lagging PF\n",
+"sin_theta_1 = sqrt(1 - (cos_theta_1)^2);\n",
+"\n",
+"// Induced voltage when the transformer is delivering rated current to unity PF load\n",
+"E1 = (V_1*cos_theta_1 + I1*R_e1) + %i*(V_1*sin_theta_1 + I1*X_e1);\n",
+"E1_m = abs(E1);//E1_m=magnitude of E1 in volt\n",
+"E1_a = atan(imag(E1) /real(E1))*180/%pi;//E1_a=phase angle of E1 in degrees\n",
+"\n",
+"VR_lag_PF = ( (E1_m - V_1) / V_1 ) * 100 ; // Transformer voltage regulation\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-19 Solution : ');\n",
+"\n",
+"printf(' \n a: High voltage impedance drop :\n    I_1*Z_e1 = V_1 = %d\n',Z_e1_drop);\n",
+"\n",
+"printf(' \n b: θ = %.f degrees \n',theta );\n",
+"printf(' \n    High voltage resistance drop :\n    I_1*R_e1 = %.2f \n',R_e1_drop);\n",
+"\n",
+"printf(' \n c: High voltage reactance drop :\n    I_1*X_e1 = %.2f \n',X_e1_drop);\n",
+"\n",
+"printf(' \n d: At unity PF,\n    E_2 in volt = ');disp(E_1);\n",
+"printf(' \n    E_2 = %.2f <%.2f V \n ',E_1_m , E_1_a);\n",
+"printf(' \n    Voltage regulation at unity PF :\n    VR = %.2f percent ',VR_unity_PF );\n",
+"\n",
+"printf(' \n\n e: At 0.7 lagging PF, \n    E_2 in volt = ');disp(E1);\n",
+"printf(' \n    E_2 = %.2f <%.2f V \n ',E1_m , E1_a);\n",
+"printf(' \n    Voltage regulation at 0.7 lagging PF :\n    VR = %.2f percent ',VR_lag_PF );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.1: stepup_stepdown_alpha_I1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data for Step -down transformer\n",
+"N_1 = 500 ; // Number of turns in the primary\n",
+"N_2 = 100 ; // Number of turns in the secondary\n",
+"I_2 = 12 ; // Load (Secondary) current in A\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"alpha = N_1 / N_2 ; // Transformation ratio\n",
+"\n",
+"// case b\n",
+"I_1 = I_2 / alpha ; // Load component of primary current in A\n",
+"\n",
+"// case c\n",
+"// sunscript c for alpha indicates case c\n",
+"// For step up transformer , using above given data\n",
+"N1 = 100 ; // Number of turns in the primary\n",
+"N2 = 500 ; // Number of turns in the secondary\n",
+"alpha_c = N1 / N2 ; // Transformation ratio\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-1 Solution : ');\n",
+"\n",
+"printf(' \n a: Transformation ratio(step-down transformer) :\n    α = %d\n',alpha);\n",
+"\n",
+"printf(' \n b: Load component of primary current : \n    I_1 = %.1f A \n',I_1);\n",
+"\n",
+"printf(' \n c: Transformation ratio(step-up transformer) :\n    α = %.1f',alpha_c);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.20: Re1_Re1_r2_its_drop_Pc.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-20\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 500 ; // kVA rating of the step-down transformer\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 208 ; // Secondary voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"\n",
+"// SC-test data\n",
+"P_sc = 8200 ; // wattmeter reading in W\n",
+"I_sc = 217.4 ; // Short circuit current in A\n",
+"V_sc = 95 ; // Short circuit voltage in V\n",
+"\n",
+"// OC-test data\n",
+"P_oc = 1800 ; // wattmeter reading in W\n",
+"I_oc = 85 ; // Open circuit current in A\n",
+"V_oc = 208 ; // Open circuit voltage in V\n",
+"\n",
+"// Calculations\n",
+"alpha = V_1 / V_2 ; // Transformation ratio\n",
+"// case a\n",
+"P = P_sc ; // wattmeter reading in W\n",
+"I1 = I_sc ; // Short circuit current in A\n",
+"R_e1 = P / (I1)^2 ; // Equivalent resistance w.r.t HV side in ohm\n",
+"R_e2 = R_e1 / (alpha)^2 // Equivalent resistance referred to LV side in ohm\n",
+"\n",
+"// case b\n",
+"r_2 = R_e2 / 2 ; // Resistance of low-voltage side in ohm\n",
+"\n",
+"// case c\n",
+"I_m = I_oc ; // Open circuit current in A\n",
+"P_cu = (I_m)^2 * r_2 ; // Transformer copper loss of the LV side wdg during OC-test in W\n",
+"\n",
+"// case d\n",
+"P_c = P_oc - P_cu ; // Transformer core loss in W\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-20 Solution : ');\n",
+"\n",
+"printf(' \n a: Equivalent resistance w.r.t HV side :\n    R_e1 = %.4f Ω\n',R_e1);\n",
+"printf(' \n    Equivalent resistance w.r.t LV side :\n    R_e2 = %f Ω = %.3f mΩ \n',R_e2,R_e2*1000);\n",
+"\n",
+"printf(' \n b: Resistance of LV side :\n    r_2 = %f Ω = %.2f mΩ\n',r_2,r_2*1000);\n",
+"\n",
+"printf(' \n c: Transformer copper loss of the LV side wdg during OC-test : ');\n",
+"printf(' \n    (I_m)^2 * r_2 = %f W \n',P_cu);\n",
+"\n",
+"printf(' \n d: Transformer core loss :\n    P_c = %f W \n ',P_c);\n",
+"\n",
+"printf(' \n e: Yes.The error is approximately 5/%d = 0.278 percent,which is',P_oc);\n",
+"printf(' \n    within the error produced by the instruments used in the test.');\n",
+"printf(' \n    We may assume that the core loss is %d W.',P_oc);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.21: tabulate_I2_efficiencies.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-21\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data(from Ex.14-18)\n",
+"V_sc = 50 ; // Short circuit voltage in volt\n",
+"V_1 = 2300 ; // Rated primary voltage in volt\n",
+"\n",
+"\n",
+"// Preliminary data before tabulating\n",
+"\n",
+"// from ex.14-20\n",
+"P_c = 1.8 ; // core losses in kW\n",
+"P_k = 1.8 ; // fixed losses in kW \n",
+"P_cu_rated = 8.2 ; // Rated copper loss in kW\n",
+"\n",
+"// given rating\n",
+"kVA = 500 ; // Power rating in kVA\n",
+"PF = 1 ; // power factor\n",
+"P_o = kVA * PF ; // full-load output at unity PF in kW\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"LF1 = 1/4 ; // Load fraction\n",
+"LF2 = 1/2 ; // Load fraction\n",
+"LF3 = 3/4 ; // Load fraction\n",
+"LF4 = 5/4 ; // Load fraction\n",
+"P_cu_fl = 8.2 ; // Equivalent copper loss at full-load slip\n",
+"P_cu_LF1 = (LF1)^2 * P_cu_fl ; // Equivalent copper loss at 1/4 rated load\n",
+"P_cu_LF2 = (LF2)^2 * P_cu_fl ; // Equivalent copper loss at 1/2 rated load\n",
+"P_cu_LF3 = (LF3)^2 * P_cu_fl ; // Equivalent copper loss at 3/4 rated load\n",
+"P_cu_LF4 = (LF4)^2 * P_cu_fl ; // Equivalent copper loss at 5/4 rated load\n",
+"\n",
+"P_L_1 = P_c + P_cu_LF1 ; // Total losses in kW at 1/4 rated load\n",
+"P_L_2 = P_c + P_cu_LF2 ; // Total losses in kW at 1/2 rated load\n",
+"P_L_3 = P_c + P_cu_LF3 ; // Total losses in kW at 3/4 rated load\n",
+"P_L_fl = P_c + P_cu_fl ; // Total losses in kW at rated load\n",
+"P_L_4 = P_c + P_cu_LF4 ; // Total losses in kW at 5/4 rated load\n",
+"\n",
+"P_o_1 = P_o*LF1 ; // Total output in kW at 1/4 rated load\n",
+"P_o_2 = P_o*LF2 ; // Total output in kW at 1/2 rated load\n",
+"P_o_3 = P_o*LF3 ; // Total output in kW at 3/4 rated load\n",
+"P_o_fl = P_o ; // Total output in kW at rated load\n",
+"P_o_4 = P_o*LF4 ; // Total output in kW at 5/4 rated load\n",
+"\n",
+"P_in_1 = P_L_1 + P_o_1 ; // Total input in kW at 1/4 rated load\n",
+"P_in_2 = P_L_2 + P_o_2 ; // Total input in kW at 1/2 rated load\n",
+"P_in_3 = P_L_3 + P_o_3 ; // Total input in kW at 3/4 rated load\n",
+"P_in_fl = P_L_fl + P_o_fl ; // Total input in kW at rated load\n",
+"P_in_4 = P_L_4 + P_o_4 ; // Total input in kW at 5/4 rated load\n",
+"\n",
+"eta_1 = (P_o_1/P_in_1)*100 ; // Efficiency at 1/4 rated load\n",
+"eta_2 = (P_o_2/P_in_2)*100 ; // Efficiency at 1/2 rated load\n",
+"eta_3 = (P_o_3/P_in_3)*100 ; // Efficiency at 3/4 rated load\n",
+"eta_fl = (P_o_fl/P_in_fl)*100 ; // Efficiency at rated load\n",
+"eta_4 = (P_o_4/P_in_4)*100 ; // Efficiency at 5/4 rated load\n",
+"\n",
+"\n",
+"// case b\n",
+"PF_b = 0.8 ; // 0.8 PF lagging\n",
+"Po_1 = P_o*LF1*PF_b ; // Total output in kW at 1/4 rated load\n",
+"Po_2 = P_o*LF2*PF_b ; // Total output in kW at 1/2 rated load\n",
+"Po_3 = P_o*LF3*PF_b ; // Total output in kW at 3/4 rated load\n",
+"Po_fl = P_o*PF_b ; // Total output in kW at rated load\n",
+"Po_4 = P_o*LF4*PF_b ; // Total output in kW at 5/4 rated load\n",
+"\n",
+"Pin_1 = P_L_1 + Po_1 ; // Total input in kW at 1/4 rated load\n",
+"Pin_2 = P_L_2 + Po_2 ; // Total input in kW at 1/2 rated load\n",
+"Pin_3 = P_L_3 + Po_3 ; // Total input in kW at 3/4 rated load\n",
+"Pin_fl = P_L_fl + Po_fl ; // Total input in kW at rated load\n",
+"Pin_4 = P_L_4 + Po_4 ; // Total input in kW at 5/4 rated load\n",
+"\n",
+"eta1 = (Po_1/Pin_1)*100 ; // Efficiency at 1/4 rated load\n",
+"eta2 = (Po_2/Pin_2)*100 ; // Efficiency at 1/2 rated load\n",
+"eta3 = (Po_3/Pin_3)*100 ; // Efficiency at 3/4 rated load\n",
+"etafl = (Po_fl/Pin_fl)*100 ; // Efficiency at rated load\n",
+"eta4 = (Po_4/Pin_4)*100 ; // Efficiency at 5/4 rated load\n",
+"\n",
+"// case c\n",
+"R_e2 = 1.417e-3 ; // Equivalent resistance in ohm referred to LV side\n",
+"Pc = 1800 ; // Core losses in W\n",
+"I_2 = sqrt(Pc/R_e2); // Load current in A for max.efficiency invariant of LF\n",
+"\n",
+"// case d\n",
+"V = 208 ; // Voltage rating in volt\n",
+"I_2_rated = (kVA*1000) / V ; // Rated secondary current in A\n",
+"LF_max = I_2 / I_2_rated ; // Load fraction for max.efficiency\n",
+"\n",
+"// case e\n",
+"// subscript e for eta_max indicates case e\n",
+"cos_theta = 1;\n",
+"V_2 = V ; // secondary voltage in volt\n",
+"Pc = 1800 ; // core loss in W\n",
+"// max.efficiency for unity PF\n",
+"eta_max_e = (V_2*I_2*cos_theta) / ((V_2*I_2*cos_theta) + (Pc + I_2^2*R_e2)) * 100\n",
+"\n",
+"// case f\n",
+"// subscript f for eta_max indicates case e\n",
+"cos_theta2 = 0.8;\n",
+"// max.efficiency for 0.8 lagging PF\n",
+"eta_max_f = (V_2*I_2*cos_theta2) / ((V_2*I_2*cos_theta2) + (Pc + I_2^2*R_e2)) * 100\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-21 Solution : ');\n",
+"\n",
+"printf(' \n a: Tabulation at unity PF : ');\n",
+"printf(' \n    __________________________________________________________________________________________________________');\n",
+"printf(' \n    L.F \t Core loss \t Copper loss \tTotal loss \t Total Output \t Total Input \t Efficiency');\n",
+"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)');\n",
+"printf(' \n    __________________________________________________________________________________________________________');\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);\n",
+"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);\n",
+"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);\n",
+"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);\n",
+"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);\n",
+"printf(' \n    __________________________________________________________________________________________________________\n\n');\n",
+"\n",
+"printf(' \n b: Tabulation at 0.8 PF lagging : ');\n",
+"printf(' \n    __________________________________________________________________________________________________________');\n",
+"printf(' \n    L.F \t Core loss \t Copper loss \tTotal loss \t Total Output \t Total Input \t Efficiency');\n",
+"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)');\n",
+"printf(' \n    __________________________________________________________________________________________________________');\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);\n",
+"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);\n",
+"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);\n",
+"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);\n",
+"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);\n",
+"printf(' \n    __________________________________________________________________________________________________________\n\n');\n",
+"\n",
+"printf(' \n c: Load current at which max.efficiency occurs :\n    I_2 = %.1f A \n',I_2);\n",
+"\n",
+"printf(' \n d: Rated load current :\n    I_2(rated) = %.1f A \n',I_2_rated);\n",
+"printf(' \n    Load fraction for η_max = %.3f(≃half rated load)\n ',LF_max);\n",
+"\n",
+"printf(' \n e: Max.efficiency for unity PF :\n    η_max = %.2f percent \n',eta_max_e);\n",
+"\n",
+"printf(' \n f: Max.efficiency for 0.8 lagging PF :\n    η_max = %.2f percent',eta_max_f);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.22: Zeqpu_V1pu_VR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-22\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"P = 20 ; // Power rating of the transformer in kVA\n",
+"// Short circuit test data\n",
+"P_sc = 250 ; // Power measured in W\n",
+"V_sc = 50 ; // Short circuit voltage in volt\n",
+"I_sc = 8.7 ; // Short circuit current in A\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_1b = V_1 ; // base voltage in volt\n",
+"Z_eq_pu = V_sc / V_1 ;\n",
+"\n",
+"funcprot(0) ; // Use this to avoid the message 'Warning : redefining function: beta '                   .\n",
+"beta = acosd(P_sc/(V_sc*I_sc)); // angle in degrees\n",
+"\n",
+"Zeq_pu = Z_eq_pu*exp(%i*(beta)*(%pi/180));\n",
+"Zeq_pu_m = abs(Zeq_pu);//Zeq_pu_m=magnitude of Zeq_pu in p.u\n",
+"Zeq_pu_a = atan(imag(Zeq_pu) /real(Zeq_pu))*180/%pi;//Zeq_pu_a=phase angle of Zeq_pu in degrees\n",
+"\n",
+"// case b\n",
+"// at unity PF\n",
+"V_1_pu = 1*exp(%i*(0)*(%pi/180)) + 1*exp(%i*(0)*(%pi/180))*Z_eq_pu*exp(%i*(beta)*(%pi/180));\n",
+"// RHS is written in exponential complex form and (%pi/180) is radians to degrees conversion factor\n",
+"V_1_pu_m = abs(V_1_pu);//V_1_pu_m=magnitude of V_1_pu in volt\n",
+"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\n",
+"\n",
+"// case c\n",
+"// at 0.7 PF lagging\n",
+"theta = acosd(0.7); // Power factor angle in degrees\n",
+"V1_pu = 1*exp(%i*(0)*(%pi/180)) + 1*exp(%i*(-theta)*(%pi/180))*Z_eq_pu*exp(%i*(beta)*(%pi/180));\n",
+"V1_pu_m = abs(V1_pu);//V1_pu_m=magnitude of V1_pu in volt\n",
+"V1_pu_a = atan(imag(V1_pu) /real(V1_pu))*180/%pi;//V1_pu_a=phase angle of V1_pu in degrees\n",
+"\n",
+"// case d\n",
+"VR_unity_PF = V_1_pu_m - 1 ; // voltage regulation at unity PF \n",
+"\n",
+"// case e\n",
+"VR_lag_PF = V1_pu_m - 1 ; // voltage regulation at 0.7 lagging PF \n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-22 Solution : ');\n",
+"\n",
+"printf(' \n a: Z_eq(pu) = %.5f p.u \n',Z_eq_pu);\n",
+"printf(' \n    β = %.f degrees \n',beta);\n",
+"printf(' \n    Z_eq(pu) <β = ');disp(Zeq_pu);\n",
+"printf(' \n    Z_eq(pu) <β = %.5f <%.f p.u \n ',Zeq_pu_m,Zeq_pu_a);\n",
+"\n",
+"printf(' \n b: |V_1(pu)| = ');disp(V_1_pu);\n",
+"printf(' \n    |V_1(pu)| = %.4f <%.2f V \n ',V_1_pu_m , V_1_pu_a );\n",
+"\n",
+"printf(' \n c: |V_1(pu)| = ');disp(V1_pu);\n",
+"printf(' \n    |V_1(pu)| = %.4f <%.2f V \n ',V1_pu_m , V1_pu_a );\n",
+"\n",
+"printf(' \n d: Voltage regulation at unity PF :\n    VR = %f ',VR_unity_PF);\n",
+"printf(' \n    VR = %.3f percent \n ',100*VR_unity_PF);\n",
+"\n",
+"printf(' \n e: Voltage regulation at 0.7 lagging PF :\n    VR = %f ',VR_lag_PF);\n",
+"printf(' \n    VR = %.2f percent \n ',100*VR_lag_PF);\n",
+"\n",
+"printf(' \n f: VRs as found by p.u method are essentially the same as those found ');\n",
+"printf(' \n    in Exs.14-17 and 14-19 using the same data, for the same transformer, ');\n",
+"printf(' \n    but with much less effort.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.23: Pcu_LF_efficiencies.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-23\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"S = 500 ; // Power rating of the transformer in kVA\n",
+"f= 60 ; // Frequency in Hz\n",
+"\n",
+"// Open circuit test data\n",
+"V_oc = 208 ; // Open circuit voltage in volt\n",
+"I_oc = 85 ; // Open circuit current in A\n",
+"P_oc = 1800 ; // Power measured in W\n",
+"\n",
+"// Short circuit test data\n",
+"V_sc = 95 ; // Short circuit voltage in volt\n",
+"I_sc = 217.4 ; // Short circuit current in A\n",
+"P_sc = 8200 ; // Power measured in W\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"S_b = S ; // Base voltage in kVA\n",
+"Psc = 8.2 ; // Power measured in kW during SC-test\n",
+"P_Cu_pu = Psc / S_b ; // per unit value of P_Cu at rated load\n",
+"\n",
+"// case b\n",
+"Poc = 1.8 ; // Power measured in kW during OC-test\n",
+"P_CL_pu = Poc / S_b ; // per unit value of P_CL at rated load\n",
+"\n",
+"// case c\n",
+"PF = 1 ; // unity Power factor\n",
+"eta_pu = PF / (PF + P_CL_pu + P_Cu_pu ) * 100 ; // Efficiency at rated load,unity PF\n",
+"\n",
+"// case d\n",
+"// subscript d for PF and eta_pu indicates case d\n",
+"PF_d = 0.8 ; // 0.8 lagging Power factor\n",
+"eta_pu_d = PF_d / (PF_d + P_CL_pu + P_Cu_pu ) * 100 ; // Efficiency at rated load,unity PF\n",
+"\n",
+"// case e\n",
+"LF = sqrt(P_CL_pu / P_Cu_pu); // Load fraction producing max.efficiency\n",
+"\n",
+"// case f\n",
+"eta_pu_max = (LF*PF) / ( (LF*PF) + 2*(P_CL_pu) ) * 100 ; // Maximum efficiency at unity PF load\n",
+"\n",
+"// case g\n",
+"eta_pu_max_g = (LF*PF_d) / ( (LF*PF_d) + 2*(P_CL_pu) ) * 100 ; // Maximum efficiency at 0.8 lagging PF load\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-23 Solution : ');\n",
+"\n",
+"printf(' \n a: Per unit copper loss at rated load :');\n",
+"printf(' \n    P_Cu(pu) = %.4f p.u = R_eq(pu)\n',P_Cu_pu);\n",
+"\n",
+"printf(' \n a: Per unit core loss at rated load :');\n",
+"printf(' \n    P_CL(pu) = %.4f p.u \n',P_CL_pu);\n",
+"\n",
+"printf(' \n c: Efficiency at rated load,unity PF :\n    η_pu = %.2f percent \n',eta_pu);\n",
+"\n",
+"printf(' \n c: Efficiency at rated load,0.8 lagging PF :\n    η_pu = %.2f percent \n',eta_pu_d);\n",
+"\n",
+"printf(' \n e: Load fraction producing max.efficiency :\n    L.F = %.3f \n ',LF );\n",
+"\n",
+"printf(' \n f: Maximum efficiency at unity PF load :\n    η_pu(max) = %.2f percent \n',eta_pu_max);\n",
+"\n",
+"printf(' \n g: Maximum efficiency at 0.8 lagging PF load :\n    η_pu(max) = %.2f percent \n',eta_pu_max_g);\n",
+"\n",
+"printf(' \n h: All efficiency values are identical to those computed in solution to Ex.14-21. \n');\n",
+"\n",
+"printf(' \n i: Per-unit method is much simpler and less subject to error than conventional method.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.24: efficiencies_at_differnt_LFs.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-24\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data(From Ex.14-23)\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"S = 500 ; // Power rating of the transformer in kVA\n",
+"f= 60 ; // Frequency in Hz\n",
+"\n",
+"// Open circuit test data\n",
+"V_oc = 208 ; // Open circuit voltage in volt\n",
+"I_oc = 85 ; // Open circuit current in A\n",
+"P_oc = 1800 ; // Power measured in W\n",
+"\n",
+"// Short circuit test data\n",
+"V_sc = 95 ; // Short circuit voltage in volt\n",
+"I_sc = 217.4 ; // Short circuit current in A\n",
+"P_sc = 8200 ; // Power measured in W\n",
+"\n",
+"// Calculations\n",
+"// Preliminary calculations\n",
+"S_b = S ; // Base voltage in kVA\n",
+"Psc = 8.2 ; // Power measured in kW during SC-test\n",
+"P_Cu_pu = Psc / S_b ; // per unit value of P_Cu at rated load\n",
+"\n",
+"Poc = 1.8 ; // Power measured in kW during OC-test\n",
+"P_CL_pu = Poc / S_b ; // per unit value of P_CL at rated load\n",
+"\n",
+"// case a\n",
+"LF1 = 3/4 ; // Load fraction of rated load \n",
+"PF1 = 1 ; // unity Power factor\n",
+"eta_pu_LF1 = (LF1*PF1) / ((LF1*PF1) + P_CL_pu + (LF1)^2*P_Cu_pu ) * 100 ; // Efficiency at rated load,unity PF\n",
+"\n",
+"// case b\n",
+"LF2 = 1/4 ; // Load fraction of rated load \n",
+"PF2 = 0.8 ; // 0.8 lagging PF\n",
+"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\n",
+"\n",
+"// case c\n",
+"LF3 = 5/4 ; // Load fraction of rated load \n",
+"PF3 = 0.8 ; // 0.8 leading PF\n",
+"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\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-24 Solution : ');\n",
+"\n",
+"printf(' \n    Efficiency(pu) :\n ');\n",
+"printf(' \n a: η_pu at %.2f rated-load = %.2f percent \n',LF1,eta_pu_LF1);\n",
+"\n",
+"printf(' \n b: η_pu at %.2f rated-load = %.2f percent \n',LF2,eta_pu_LF2);\n",
+"\n",
+"printf(' \n c: η_pu at %.2f rated-load = %.2f percent \n',LF3,eta_pu_LF3);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.25: Zpu2_St_S2_S1_LF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-25\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA_1 = 500 ; // Power rating of the transformer 1 in kVA\n",
+"R_1_pu = 0.01 ; // per-unit value of resistance of the transformer 1\n",
+"X_1_pu = 0.05 ; // per-unit value of reactance of the transformer 1\n",
+"Z_1_pu = R_1_pu + %i*X_1_pu ; //per-unit value of impedance of the transformer 1\n",
+"\n",
+"PF = 0.8 ; // lagging power factor\n",
+"V_2 = 400 ; // Secondary voltage in volt\n",
+"S_load = 750 ; // Increased system load in kVA\n",
+"\n",
+"kVA_2 = 250 ; // Power rating of the transformer  2 in kVA\n",
+"R_pu_2 = 0.015 ; // per-unit value of resistance of the transformer 2\n",
+"X_pu_2 = 0.04 ; // per-unit value of reactance of the transformer 2\n",
+"\n",
+"// smaller transformer secondary voltageis same as larger transformer\n",
+"\n",
+"// Calculations\n",
+"// Preliminary calculations\n",
+"Z_pu_1 = R_pu_2 + %i*X_pu_2 ; // New transformer p.u. impedance\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_b1 = 400 ; // base voltage in volt\n",
+"V_b2 = 400 ; // base voltage in volt\n",
+"Z_pu_2 = (kVA_1/kVA_2)*(V_b1/V_b2)^2 * (Z_pu_1); // New transformer p.u impedance\n",
+"Z_2_pu = Z_pu_2 ; //New transformer p.u impedance\n",
+"\n",
+"// case b\n",
+"cos_theta = PF ; // Power factor\n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"S_t_conjugate = (kVA_1 + kVA_2)*(cos_theta + %i*sin_theta); // kVA of total load\n",
+"\n",
+"// case c\n",
+"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\n",
+"S_2_conjugate_m = abs(S_2_conjugate);//S_2_conjugate_m=magnitude of S_2_conjugate in kVA\n",
+"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\n",
+"\n",
+"// case d\n",
+"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\n",
+"S_1_conjugate_m = abs(S_1_conjugate);//S_1_conjugate_m=magnitude of S_1_conjugate in kVA\n",
+"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\n",
+"\n",
+"// case e\n",
+"S_1 = S_1_conjugate_m ; \n",
+"S_b1 = kVA_1 ; // base power in kVA of trancsformer 1\n",
+"LF1 = (S_1 / S_b1)*100 ; // Load fraction of the original transformer in percent\n",
+"\n",
+"// case f\n",
+"S_2 = S_2_conjugate_m ; \n",
+"S_b2 = kVA_2 ; // base power in kVA of trancsformer 2\n",
+"LF2 = (S_2 / S_b2)*100 ; // Load fraction of the original transformer in percent\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-25 Solution : ');\n",
+"\n",
+"printf(' \n a: New transformer p.u impedance :\n    Z_p.u.2 in p.u = ');disp(Z_pu_2);\n",
+"\n",
+"printf(' \n b: kVA of total load :\n    S*_t in kVA = ');disp(S_t_conjugate);\n",
+"\n",
+"printf(' \n c: Portion of load carried by the smaller transformer in kVA :');\n",
+"printf(' \n    S*_2 in kVA =  ');disp(S_2_conjugate);\n",
+"printf(' \n    S*_2 = %.1f <%.2f kVA (inductive load)\n',S_2_conjugate_m,S_2_conjugate_a);\n",
+"\n",
+"printf(' \n d: Portion of load carried by the original transformer in kVA :');\n",
+"printf(' \n    S*_2 in kVA = ');disp(S_1_conjugate);\n",
+"printf(' \n    S*_2 = %.1f <%.2f kVA (inductive load)\n',S_1_conjugate_m,S_1_conjugate_a);\n",
+"\n",
+"printf(' \n e: Load fraction of the original transformer :\n    L.F.1 = %.1f percent\n',LF1);\n",
+"\n",
+"printf(' \n f: Load fraction of the original transformer :\n    L.F.2 = %.1f percent\n',LF2);\n",
+"\n",
+"printf(' \n g: Yes. Reduce the no-load voltage of the new transformer to some value ');\n",
+"printf(' \n    below that of its present value so that its share of the load is reduced.');\n",
+"\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.26: Vb_Ib_Zb_Z1_Z2_I1_I2_E1_E2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-26\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data(From Ex.14-25)\n",
+"kVA_1 = 500 ; // Power rating of the transformer 1 in kVA\n",
+"R_1_pu = 0.01 ; // per-unit value of resistance of the transformer 1\n",
+"X_1_pu = 0.05 ; // per-unit value of reactance of the transformer 1\n",
+"Z_1_pu = R_1_pu + %i*X_1_pu ; //per-unit value of impedance of the transformer 1\n",
+"\n",
+"PF = 0.8 ; // lagging power factor\n",
+"V = 400 ; // Secondary voltage in volt\n",
+"S_load = 750 ; // Increased system load in kVA\n",
+"\n",
+"kVA_2 = 250 ; // Power rating of the transformer  2 in kVA\n",
+"R_pu_2 = 0.015 ; // per-unit value of resistance of the transformer 2\n",
+"X_pu_2 = 0.04 ; // per-unit value of reactance of the transformer 2\n",
+"\n",
+"// smaller transformer secondary voltageis same as larger transformer\n",
+"\n",
+"// Calculations\n",
+"// Preliminary calculations\n",
+"Z_pu_1 = R_pu_2 + %i*X_pu_2 ; // New transformer p.u. impedance\n",
+"\n",
+"// case a\n",
+"V_b = V ; // (given)\n",
+"\n",
+"//case b\n",
+"S_b =500*1000 ; // base power in VA\n",
+"I_b = S_b / V_b ; // base current in A\n",
+"\n",
+"// case c\n",
+"Z_b = V^2/S_b ; // Base impedance in ohm\n",
+"\n",
+"// case d\n",
+"Z_1 = Z_b * Z_1_pu * 1000 ; // Actual impedance of larger transformer in milli-ohm\n",
+"Z_1_m = abs(Z_1);//Z_1_m=magnitude of Z_1 in ohm\n",
+"Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees\n",
+"\n",
+"// case e\n",
+"V_b1 = V_b ; // base voltage in volt\n",
+"V_b2 = V_b ; // base voltage in volt\n",
+"Z_pu_2 = (kVA_1/kVA_2)*(V_b1/V_b2)^2 * (Z_pu_1); // New transformer p.u impedance\n",
+"Z_2_pu = Z_pu_2 ; //New transformer p.u impedance\n",
+"\n",
+"Z_2 = Z_b * Z_2_pu*1000 ; // Actual impedance of smaller transformer in milli-ohm\n",
+"Z_2_m = abs(Z_2);//Z_2_m=magnitude of Z_2 in ohm\n",
+"Z_2_a = atan(imag(Z_2) /real(Z_2))*180/%pi;//Z_2_a=phase angle of Z_2 in degrees\n",
+"\n",
+"// case f\n",
+"cos_theta = 0.8 ; // Power factor\n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"S_T = (kVA_1 + kVA_2)*(cos_theta - %i*sin_theta); // kVA of total load\n",
+"\n",
+"I_T = S_T*1000 / V_b ; // Total current in A\n",
+"\n",
+"I_1 = I_T*(Z_2/(Z_1 + Z_2)); // Actual current delivered by larger transformer in A\n",
+"I_1_m = abs(I_1);//I_1_m=magnitude of I_1 in A\n",
+"I_1_a = atan(imag(I_1) /real(I_1))*180/%pi;//I_1_a=phase angle of I_1 in degrees\n",
+"\n",
+"// case g\n",
+"I_2 = I_T*(Z_1/(Z_1 + Z_2)); // Actual current delivered by larger transformer in A\n",
+"I_2_m = abs(I_2);//I_2_m=magnitude of I_2 in A\n",
+"I_2_a = atan(imag(I_2) /real(I_2))*180/%pi;//I_2_a=phase angle of I_2 in degrees\n",
+"\n",
+"// case h\n",
+"Z1 = Z_1/1000 ; // Z_1 in ohm\n",
+"E_1 = I_1*Z1 + V_b ; // No-load voltage of larger Tr. in volt\n",
+"E_1_m = abs(E_1);//E_1_m=magnitude of E_1 in volt\n",
+"E_1_a = atan(imag(E_1) /real(E_1))*180/%pi;//E_1_a=phase angle of E_1 in degrees\n",
+"\n",
+"\n",
+"// case i\n",
+"Z2 = Z_2/1000 ; // Z_2 in ohm\n",
+"E_2 = I_2*Z2 + V_b ; // No-load voltage of smaller Tr. in volt\n",
+"E_2_m = abs(E_2);//E_2_m=magnitude of E_2 in volt\n",
+"E_2_a = atan(imag(E_2) /real(E_2))*180/%pi;//E_2_a=phase angle of E_2 in degrees\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-26 Solution : ');\n",
+"\n",
+"printf(' \n a: Base voltage :\n    V_b = %d <0 V (given)\n',V_b);\n",
+"\n",
+"printf(' \n b: Base current :\n    I_b = %.2f A \n',I_b);\n",
+"\n",
+"printf(' \n c: Base impedance :\n    Z_b = %.2f ohm\n',Z_b);\n",
+"\n",
+"printf(' \n d: Actual impedance of larger transformer :\n    Z_1 in mΩ = \n');disp(Z_1);\n",
+"printf(' \n    Z_1 = %.2f <%.2f mΩ \n ',Z_1_m,Z_1_a);\n",
+"\n",
+"printf(' \n e: Actual impedance of smaller transformer :\n    Z_1 in mΩ = \n');disp(Z_2);\n",
+"printf(' \n    Z_1 = %.2f <%.2f mΩ \n ',Z_2_m,Z_2_a);\n",
+"\n",
+"printf(' \n f: Actual current delivered by larger transformer :\n    I_1 in A = ');disp(I_1);\n",
+"printf(' \n    I_1 = %.2f <%.2f A \n ',I_1_m,I_1_a);\n",
+"\n",
+"printf(' \n g: Actual current delivered by smaller transformer :\n    I_2 in A = ');disp(I_2);\n",
+"printf(' \n    I_1 = %.2f <%.2f A \n ',I_2_m,I_2_a);\n",
+"\n",
+"printf(' \n h: No-load voltage of larger Tr :\n    E_1 in volt = ');disp(E_1);\n",
+"printf(' \n    E_1 = %.2f <%.2f V \n ',E_1_m,E_1_a);\n",
+"\n",
+"printf(' \n i: No-load voltage of smaller Tr :\n    E_2 in volt = ');disp(E_2);\n",
+"printf(' \n    E_1 = %.2f <%.2f V \n ',E_2_m,E_2_a);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.27: RL_ZbL_ZLpu_Z2pu_Z1pu_IbL_ILpu_VRpu_VSpu_VS_VxVxpu.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-27\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// From diagram in fig.14-23a\n",
+"P_L = 14400 ; // Load output power in W\n",
+"V_L = 120 ; // Load voltage in volt\n",
+"V_b1 = 120 ; // base voltage at point 1 in volt\n",
+"V_b2 = 600 ; // base voltage at point 2 in volt\n",
+"V_b3 = 120 ; // base voltage at point 3 in volt\n",
+"S_b3 = 14.4 ; // base power in kVA\n",
+"X_2 = %i*0.25 ; // reactance in p.u\n",
+"X_1 = %i*0.2 ; // reactance in p.u\n",
+"I_L = 120 ; // Load current in A\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"R_L = P_L / (V_L^2); // Resistance of the load in ohm\n",
+"\n",
+"// case b\n",
+"Z_bL = (V_b3^2)/(S_b3*1000); // Base impedance in ohm\n",
+"\n",
+"// case c\n",
+"Z_L_pu = R_L / Z_bL ; // per unit load impedance \n",
+"\n",
+"// case d\n",
+"Z_2_pu = X_2 ; // per unit impedance of Tr.2\n",
+"\n",
+"// case e\n",
+"Z_1_pu = X_1 ; // per unit impedance of Tr.1\n",
+"\n",
+"// case g\n",
+"I_bL = (S_b3*1000)/V_b3 ; // Base current in load in A\n",
+"\n",
+"// case h\n",
+"I_L_pu = I_L / I_bL ; // per unit load current\n",
+"\n",
+"// case i\n",
+"V_R_pu = I_L_pu * Z_L_pu ; // per unit voltage across load \n",
+"\n",
+"// case j\n",
+"I_S_pu = I_L_pu ; //per unit current of source\n",
+"Z_T_pu = Z_L_pu + Z_1_pu + Z_2_pu ; // Total p.u impedance\n",
+"V_S_pu = I_S_pu * Z_T_pu ; // per unit voltage of source\n",
+"V_S_pu_m = abs(V_S_pu);//V_S_pu_m=magnitude of V_S_pu in p.u\n",
+"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\n",
+"\n",
+"// case k\n",
+"V_S = V_S_pu * V_b1 ; // Actual voltage across source in volt\n",
+"V_S_m = abs(V_S);//V_S_m=magnitude of V_S in volt\n",
+"V_S_a = atan(imag(V_S) /real(V_S))*180/%pi;//V_S_a=phase angle of V_S in degrees\n",
+"\n",
+"\n",
+"// case l\n",
+"I_x_pu = I_L_pu ; // p.u current at point x\n",
+"Z_x_pu = Z_L_pu + Z_2_pu ; // p.u impedance at point x \n",
+"V_x_pu = I_x_pu * Z_x_pu ; // p.u voltage at point x\n",
+"\n",
+"// case m\n",
+"V_x = V_x_pu * V_b2 ; // Actual voltage at point x in volt\n",
+"V_x_m = abs(V_x);//V_x_m=magnitude of V_x in volt\n",
+"V_x_a = atan(imag(V_x) /real(V_x))*180/%pi;//V_x_a=phase angle of V_x in degrees\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-27 Solution : ');\n",
+"\n",
+"printf(' \n a: Resistance of the load :\n    R_L = %d Ω \n',R_L);\n",
+"\n",
+"printf(' \n b: Base impedance :\n    Z_bL = %d Ω \n',Z_bL);\n",
+"\n",
+"printf(' \n c: per unit load impedance :\n    Z_L(pu)  = ');disp(Z_L_pu);\n",
+"\n",
+"printf(' \n d: per unit impedance of Tr.2 :\n    Z_2(pu) = ');disp(Z_2_pu);\n",
+"\n",
+"printf(' \n e: per unit impedance of Tr.1 :\n    Z_1(pu)  = ');disp(Z_1_pu);\n",
+"\n",
+"printf(' \n f: See Fig.14-23b \n');\n",
+"\n",
+"printf(' \n g: Base current in load :\n    I_bL = %d A (resistive)\n',I_bL);\n",
+"\n",
+"printf(' \n h: per unit load current :\n    I_L_pu = ');disp(I_L_pu);\n",
+"\n",
+"printf(' \n i: per unit voltage across load :\n    V_R_pu');disp(V_R_pu);\n",
+"\n",
+"printf(' \n j: per unit voltage of source :\n    V_S_pu = ');disp(V_S_pu);\n",
+"printf(' \n    V_S_pu = %.3f <%.2f p.u \n',V_S_pu_m,V_S_pu_a);\n",
+"\n",
+"printf(' \n k: Actual voltage across source :\n    V_S in volt = ');disp(V_S);\n",
+"printf(' \n    V_S = %.1f <%.2f V \n',V_S_m,V_S_a);\n",
+"\n",
+"printf(' \n l: p.u voltage at point x :\n    V_x(pu) = ');disp(V_x_pu);\n",
+"\n",
+"printf(' \n m: Actual voltage at point x  :\n    V_x in volt = ');disp(V_x);\n",
+"printf(' \n    V_S = %.1f <%.2f V \n',V_x_m,V_x_a);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.28: ZT1_ZT2_Zbline3_Zlinepu_VLpu_IbL_IL_ILpu_VSpu_VS.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-28\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// From diagram in fig.14-24a\n",
+"V_1 = 11 ; // Tr.1 voltage in kV\n",
+"V_b1 = 11 ; // Base Tr.1 voltage in kV\n",
+"S_1 = 50 ; // KVA rating of power for Tr.1\n",
+"S_2 = 100 ; // KVA rating of power for Tr.2\n",
+"Z_1_pu = %i*0.1 ; // per unit impedance of Tr.1\n",
+"Z_2_pu = %i*0.1 ; // per unit impedance of Tr.2\n",
+"V_b2 = 55 ; // Base Tr.2 voltage in kV\n",
+"S_b = 100 ; // base power in kVA\n",
+"PF = 0.8 ; // power factor of the Tr.s\n",
+"\n",
+"Z_line = %i*200 ; // line impedance in ohm\n",
+"\n",
+"V_L = 10 ;// Load voltage in kV\n",
+"V_Lb3 = 11 ; // base line voltage at point 3\n",
+"\n",
+"V_b3 = 11 ; // line voltage at point 3\n",
+"\n",
+"P_L = 50 ; // Power rating of each Tr.s in kW\n",
+"cos_theta_L = 0.8  ; // PF operation of each Tr.s\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"Z_T1 = Z_1_pu * (V_1/V_b1)^2 * (S_2/S_1); // p.u impedance of Tr.1\n",
+"\n",
+"// case b\n",
+"Z_T2 = Z_2_pu * (V_1/V_b3)^2 * (S_2/S_1); // p.u impedance of Tr.1\n",
+"\n",
+"// case c\n",
+"V_b = 55 ; // base voltage in volt\n",
+"Z_b_line = (V_b^2)/S_b * 1000 ; // base line impedance in ohm\n",
+"Z_line_pu = Z_line / Z_b_line ; // p.u impedance of the transmission line\n",
+"\n",
+"// case d\n",
+"V_L_pu = V_L / V_Lb3 ; // p.u voltage across load\n",
+"\n",
+"// case e\n",
+"//See Fig.14-24b\n",
+"\n",
+"// case f\n",
+"I_bL = S_b / V_b3 ; // base current in load in A\n",
+"\n",
+"// case g\n",
+"VL = 11 ; // load voltage in kV\n",
+"cos_theta_L = 0.8 ; // power factor\n",
+"I_L = P_L / (VL*cos_theta_L);\n",
+"I_L_pu = I_L / I_bL ; // p.u load current\n",
+"theta = acosd(0.8);\n",
+"I_Lpu = I_L_pu*(cosd(theta) - %i*sind(theta)) ;// p.u current in complex form\n",
+"\n",
+"// case h\n",
+"Z_series_pu = Z_T1 + Z_line_pu + Z_T2 ; // p.u series impedance os the transmission line\n",
+"V_S_pu = I_Lpu * Z_series_pu + V_L_pu ; // p.u source voltage\n",
+"V_S_pu_m = abs(V_S_pu);//V_S_pu_m=magnitude of V_S_pu in p.u\n",
+"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\n",
+"\n",
+"// case i\n",
+"V_S = V_S_pu_m * V_b1 ; // Actual value of source voltage in kV\n",
+"V_source = V_S*exp(%i*(V_S_pu_a)*(%pi/180)); // V_S in exponential form\n",
+"V_source_m = abs(V_source);//V_source_m=magnitude of V_source in p.u\n",
+"V_source_a = atan(imag(V_source) /real(V_source))*180/%pi;//V_source_a=phase angle of V_source in degrees\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-28 Solution : ');\n",
+"\n",
+"printf(' \n a: p.u impedance of Tr.1 :\n    Z_T1 = ');disp(Z_T1);\n",
+"\n",
+"printf(' \n b: p.u impedance of Tr.2 :\n    Z_T2 = ');disp(Z_T2);\n",
+"\n",
+"printf(' \n c: base line impedance in ohm :\n    Z_b(line) = %d ohm \n',Z_b_line);\n",
+"printf(' \n    p.u impedance of the transmission line :\n    Z(line)_pu = ');disp(Z_line_pu);\n",
+"\n",
+"printf(' \n d: p.u voltage across load :\n    V_L_pu = ');disp(V_L_pu);\n",
+"\n",
+"printf(' \n e: See Fig.14-24b \n');\n",
+"\n",
+"printf(' \n f: base current in load :\n    I_bL = %.3f A \n',I_bL);\n",
+"\n",
+"printf(' \n g: Load current :\n    I_L = %f A \n',I_L);\n",
+"printf(' \n    p.u load current:\n    I_L_pu = %.3f at %.1f PF lagging \n',I_L_pu,PF);\n",
+"printf(' \n    p.u current in complex form :\n    I_L_pu = ');disp(I_Lpu);\n",
+"\n",
+"printf(' \n h: per unit voltage of source :\n    V_S_pu = ');disp(V_S_pu);\n",
+"printf(' \n    V_S_pu = %.3f <%.2f p.u \n',V_S_pu_m,V_S_pu_a);\n",
+"\n",
+"printf(' \n i: Actual voltage across source :\n    V_S in kV = ');disp(V_source);\n",
+"printf(' \n    V_S = %.1f <%.2f kV \n',V_source_m,V_source_a);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.29: Z1pu_Z2pu_Vbline_Zlinepu_ZMs.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-29\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// From diagram in fig.14-25a\n",
+"Z_pu_1 = %i*0.1 ; //  p.u impedance \n",
+"MVA_2 = 80 ; // MVA rating os system 2\n",
+"MVA_1 = 100 ; // MVA rating of Tr.s 1 and 2\n",
+"V_2 = 30 ; // voltage in KV\n",
+"V_1 = 32 ; // voltage in KV\n",
+"\n",
+"Z_pu_2 = %i*0.15 ; //  p.u impedance \n",
+"\n",
+"V_b1 = 100 ; // base voltage of Tr.1\n",
+"\n",
+"Z_line = %i*60 ; // Line impedance \n",
+"\n",
+"MVA_M1 = 20 ; // MVA rating of motor load 1\n",
+"Z_pu_M1 = %i*0.15 ; // p.u impedance of motor load M1\n",
+"\n",
+"MVA_M2 = 35 ; // MVA rating of motor load 2\n",
+"Z_pu_M2 = %i*0.25 ; // p.u impedance of motor load M2\n",
+"\n",
+"MVA_M3 = 25 ; // MVA rating of motor load 3\n",
+"Z_pu_M3 = %i*0.2 ; // p.u impedance of motor load M3\n",
+"\n",
+"V_M = 28 ; // voltage across motor loads M1,M2,M3 in kV\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"Z_1_pu = Z_pu_1*(MVA_2/MVA_1)*(V_2/V_1)^2 ; // p.u imepedance of T1\n",
+"\n",
+"// case b\n",
+"Z_2_pu = Z_pu_2*(MVA_2/MVA_1)*(V_2/V_1)^2 ; // p.u imepedance of T2\n",
+"\n",
+"// case c\n",
+"V_b_line = V_b1*(V_1/V_2) ; // base voltage of the long-transmission line in kV\n",
+"\n",
+"// case d\n",
+"MVA_b = 80 ; // MVA rating\n",
+"V_b = V_b_line ;\n",
+"Z_line_pu = Z_line*(MVA_b/(V_b)^2); // p.u impedance of the transmission line\n",
+"\n",
+"// case e\n",
+"Z_M1_pu = Z_pu_M1 * (MVA_2/MVA_M1)*(V_M/V_1)^2 ; // p.u impedance of motor load M1\n",
+"Z_M2_pu = Z_pu_M2 * (MVA_2/MVA_M2)*(V_M/V_1)^2 ; // p.u impedance of motor load M2\n",
+"Z_M3_pu = Z_pu_M3 * (MVA_2/MVA_M3)*(V_M/V_1)^2 ; // p.u impedance of motor load M3\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-29 Solution : ');\n",
+"\n",
+"printf(' \n a: p.u imepedance of T1 :\n    Z_1(pu) = ');disp(Z_1_pu);\n",
+"\n",
+"printf(' \n b: p.u imepedance of T2 :\n    Z_2(pu) = ');disp(Z_2_pu);\n",
+"\n",
+"printf(' \n c: base voltage of the long-transmission line :\n    V_b(line) = %.1f kV \n',V_b_line);\n",
+"\n",
+"printf(' \n d: p.u impedance of the transmission line :\n    Z(line)_pu = ');disp(Z_line_pu);\n",
+"\n",
+"printf(' \n e: p.u impedance of motor load M1 :\n    Z_M1(pu) = ');disp(Z_M1_pu);\n",
+"\n",
+"printf(' \n f: p.u impedance of motor load M1 :\n    Z_M2(pu) = ');disp(Z_M2_pu);\n",
+"\n",
+"printf(' \n g: p.u impedance of motor load M1 :\n    Z_M3(pu) = ');disp(Z_M3_pu);\n",
+"\n",
+"printf(' \n h: See Fig.14-25b.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.2: turns_I1_I2_stepup_stepdown_alpha.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_h = 2300 ; // high voltage in volt\n",
+"V_l = 115 ; // low voltage in volt\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 115 ; // Secondary voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"S = 4.6 ; // kVA rating of the step-down transformer\n",
+"S_1 = S ; \n",
+"S_2 = S ;\n",
+"V_per_turn = 2.5 ; // Induced EMF per turn in V/turn\n",
+"// Ideal transformer\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"N_h = V_h / V_per_turn ; // Number of high-side turns\n",
+"N_l = V_l / V_per_turn ; // Number of low-side turns\n",
+"\n",
+"N_1 = N_h;// Number of turns in the primary\n",
+"N_2 = N_l;// Number of turns in the secondary\n",
+"\n",
+"// case b\n",
+"I_1 = S_1*1000 / V_1 ; // Rated primary current in A\n",
+"I_2 = S_2*1000 / V_2 ; // Rated secondary current in A\n",
+"\n",
+"I_h = 2 ; // Rated current in A on HV side\n",
+"I_l = 40 ; // Rated current in A on LV side\n",
+"\n",
+"// case c\n",
+"// subscript c for alpha_stepdown and alpha_stepup indicates case c\n",
+"alpha_stepdown_c = N_1 / N_2 ; // step-down transformation ratio\n",
+"alpha_stepup_c = N_l / N_h ; // step-up transformation ratio\n",
+"\n",
+"// case d\n",
+"// subscript d for alpha_stepdown and alpha_stepup indicates case d\n",
+"alpha_stepdown_d = I_2 / I_1 ; // step-down transformation ratio\n",
+"alpha_stepup_d = I_h / I_l ; // step-up transformation ratio\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-2 Solution : ');\n",
+"\n",
+"printf(' \n a: Number of high-side turns :\n    N_h = %d t = N_1 \n',N_h);\n",
+"printf(' \n    Number of low-side turns :\n    N_l = %d t = N_2\n',N_l);\n",
+"\n",
+"printf(' \n b: Rated primary current :\n    I_h = I_1 = %d A \n',I_1);\n",
+"printf(' \n    Rated secondary current :\n    I_l = I_2 = %d A\n',I_2);\n",
+"\n",
+"printf(' \n c: step-down transformation ratio :\n    α = N_1/N_2 = %d\n',alpha_stepdown_c);\n",
+"printf(' \n    step-up transformation ratio :\n    α = N_l/N_h = %.2f\n',alpha_stepup_c);\n",
+"\n",
+"printf(' \n d: step-down transformation ratio :\n    α = I_2 / I_1 = %d\n',alpha_stepdown_d);\n",
+"printf(' \n    step-up transformation ratio :\n    α = I_h / I_lh = %.2f\n',alpha_stepup_d);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.30: ST_ST_Sxformer.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-30\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// subscripts a,b,c for the current, voltages indicates respective cases a ,b ,c.\n",
+"// from fig.14-27a\n",
+"V_pa = 1000 ; // Phase voltage in volt\n",
+"I_1a = 1 ; // line current in primary in A\n",
+"V_2a = 100 ; // voltage across secondary in V\n",
+"Ic_a = 10 ; // current in lower half of auto-transformer in A\n",
+"\n",
+"// from fig.14-26b\n",
+"V_s = 100 ; // voltage in secondary wdg in V\n",
+"I_2b = 10 ; // current in secondary in A\n",
+"V_1b = 1000 ; // voltage across primary in V\n",
+"Ic_b = 1 ; // current in lower half of auto-transformer in A\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"S_T1 = (V_pa*I_1a + V_2a*I_1a)/1000 ; // Total kVA transfer in step-down mode\n",
+"\n",
+"// case b\n",
+"S_T2 = (V_s*I_2b + V_1b*I_2b)/1000 ; // Total kVA transfer in step-up mode\n",
+"\n",
+"// case c\n",
+"S_x_former_c = V_pa*I_1a/1000 ; // kVA rating of th autotransformer  in Fig.14-27a\n",
+"\n",
+"// case d\n",
+"V_1 = V_pa ;\n",
+"S_x_former_d = V_1*Ic_b/1000  ; // kVA rating of th autotransformer  in Fig.14-26b\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-30 Solution : ');\n",
+"\n",
+"printf(' \n a: Total kVA transfer in step-down mode :\n    S_T = %.1f kVA transferred \n',S_T1);\n",
+"\n",
+"printf(' \n b: Total kVA transfer in step-up mode :\n    S_T = %.1f kVA transferred \n',S_T2);\n",
+"\n",
+"printf(' \n c: kVA rating of th autotransformer in Fig.14-27a:\n    S_x-former = %d kVA \n ',S_x_former_c);\n",
+"\n",
+"printf(' \n d: kVA rating of th autotransformer in Fig.14-26b:\n    S_x-former = %d kVA \n ',S_x_former_d);\n",
+"\n",
+"printf(' \n e: Both transformers have the same kVA rating of 1 kVA since the same ');\n",
+"printf(' \n    autotransformer is used in both parts.Both transformers transform ');\n",
+"printf(' \n    a total of 1 KVA. But the step-down transformer in part(a) conducts ');\n",
+"printf(' \n    only 0.1 kVA while the step-up transformer in the part(b) conducts 10');\n",
+"printf(' \n    kVA from the primary to the secondary.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.31: Wc_tabulate_allday_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-31\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"S = 500 ; // kVA rating of distribution transformer\n",
+"// given data from ex.14-20\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 208 ; // Secondary voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"\n",
+"// SC-test data\n",
+"P_sc = 8200 ; // wattmeter reading in W\n",
+"I_sc = 217.4 ; // Short circuit current in A\n",
+"V_sc = 95 ; // Short circuit voltage in V\n",
+"\n",
+"// OC-test data\n",
+"P_oc = 1800 ; // wattmeter reading in W\n",
+"I_oc = 85 ; // Open circuit current in A\n",
+"V_oc = 208 ; // Open circuit voltage in V\n",
+"\n",
+"LF_1 = 20 ; // Load fraction in percent\n",
+"LF_2 = 40 ; // Load fraction in percent \n",
+"LF_3 = 80 ; // Load fraction in percent\n",
+"LF_fl = 100 ; // rated load in percent\n",
+"LF_4 = 125 ; // Load fraction in percent\n",
+"\n",
+"LF1 = 0.2 ; // Load fraction \n",
+"LF2 = 0.4 ; // Load fraction \n",
+"LF3 = 0.8 ; // Load fraction \n",
+"LF4 = 1.25 ; // Load fraction \n",
+"\n",
+"PF1 = 0.7 ; // power factor\n",
+"PF2 = 0.8 ; // power factor\n",
+"PF3 = 0.9 ; // power factor\n",
+"PF_fl = 1 ; // power factor\n",
+"PF4 = 0.85 ; // power factor\n",
+"\n",
+"t1 = 4 ; // period of operation in hours\n",
+"t2 = 4 ; // period of operation in hours\n",
+"t3 = 6 ; // period of operation in hours\n",
+"t_fl = 6 ; // period of operation in hours\n",
+"t4 = 2 ; // period of operation in hours\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"t = 24 ; // hrs in a day\n",
+"P_c  = P_oc ; // wattmeter reading in W (OC test)\n",
+"W_c = (P_c * t)/1000 ; // COre loss over 24 hour period\n",
+"\n",
+"// case b\n",
+"Psc = P_sc/1000 ; // wattmeter reading in W (SC test)\n",
+"P_loss_1 = (LF1^2)*Psc ; // Power loss in kW for 20% Load\n",
+"P_loss_2 = (LF2^2)*Psc ; // Power loss in kW for 40% Load\n",
+"P_loss_3 = (LF3^2)*Psc ; // Power loss in kW for 80% Load\n",
+"P_loss_fl = Psc ; // Power loss in kW for 100% Load\n",
+"P_loss_4 = (LF4^2)*Psc ; // Power loss in kW for 125% Load\n",
+"\n",
+"// energy loss in kWh\n",
+"energy_loss1 = P_loss_1 * t1 ; // Enegry loss in kWh for 20% Load\n",
+"energy_loss2 = P_loss_2 * t2 ; // Enegry loss in kWh for 40% Load\n",
+"energy_loss3 = P_loss_3 * t3 ; // Enegry loss in kWh for 80% Load\n",
+"energy_loss_fl = P_loss_fl * t_fl ; // Enegry loss in kWh for 100% Load\n",
+"energy_loss4 = P_loss_4 * t4 ; // Enegry loss in kWh for 125% Load\n",
+"\n",
+"// Total energy losses in 24hrs\n",
+"W_loss_total = energy_loss1 + energy_loss2 + energy_loss3 + energy_loss_fl + energy_loss4 ;\n",
+"\n",
+"// case c\n",
+"P_1 = LF1*S*PF1 ; // Power output for 20% load\n",
+"P_2 = LF2*S*PF2 ; // Power output for 40% load\n",
+"P_3 = LF3*S*PF3 ; // Power output for 80% load\n",
+"P_fl = S*PF_fl ; // Power output for 100% load\n",
+"P_4 = LF4*S*PF4 ; // Power output for 125% load\n",
+"\n",
+"Energy_1 = P_1*t1 ; // Energy delivered in kWh for 20%load\n",
+"Energy_2 = P_2*t2 ; // Energy delivered in kWh for 40%load\n",
+"Energy_3 = P_3*t3 ; // Energy delivered in kWh for 80%load\n",
+"Energy_fl = P_fl*t_fl ; // Energy delivered in kWh for 100%load\n",
+"Energy_4 = P_4*t4 ; // Energy delivered in kWh for 125%load\n",
+"\n",
+"// Total energy delivered in 24hrs\n",
+"W_out_total = Energy_1 + Energy_2 + Energy_3 + Energy_fl + Energy_4 ;\n",
+"\n",
+"// case d\n",
+"eta = W_out_total / (W_out_total + W_c + W_loss_total) * 100 ; // All-day efficiency\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-31 Solution : ');\n",
+"\n",
+"printf(' \n a: Total energy core loss for 24hrs, including 2hours at no-load,');\n",
+"printf(' \n    W_c = %.1f kWh \n ',W_c);\n",
+"\n",
+"printf(' \n b: From SC test, equivalent copper loss at rated load = %.1f kW,  ',Psc);\n",
+"printf(' \n    and the various energy losses during the 24 hr period are tabulated as :\n');\n",
+"\n",
+"printf(' \n    _____________________________________________________________________________________');\n",
+"printf(' \n    Percent Rated load \t Power loss(kW) \t Time period(hours) \t Energy loss(kWh)');\n",
+"printf(' \n    _____________________________________________________________________________________');\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);\n",
+"printf(' \n\t\t%d \t  %f \t\t\t %d \t\t\t %.2f \n ',LF_2,P_loss_2,t2,energy_loss2);\n",
+"printf(' \n\t\t%d \t  %f \t\t\t %d \t\t\t %.2f \n ',LF_3,P_loss_3,t3,energy_loss3);\n",
+"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);\n",
+"printf(' \n\t\t%d \t  %f \t\t\t %d \t\t\t %.2f \n ',LF_4,P_loss_4,t4,energy_loss4);\n",
+"printf(' \n    _____________________________________________________________________________________');\n",
+"printf(' \n    Total energy load losses over 24hour period (excluding 2hrs at no-load)  =  %.2f ',W_loss_total);\n",
+"printf(' \n    _____________________________________________________________________________________\n\n');\n",
+"\n",
+"printf(' \n c: Total energy output over the 24 hour period is tabulated as : \n');\n",
+"\n",
+"printf(' \n    _____________________________________________________________________________________');\n",
+"printf(' \n    Percent Rated load \t PF \t kW \t Time period(hours) \t Energy delivered(kWh)');\n",
+"printf(' \n    _____________________________________________________________________________________');\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);\n",
+"printf(' \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ',LF_2,PF2,P_2,t2,Energy_2);\n",
+"printf(' \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ',LF_3,PF3,P_3,t3,Energy_3);\n",
+"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);\n",
+"printf(' \n\t\t%d \t %.1f \t %.f \t\t %d \t\t\t %d ',LF_4,PF4,P_4,t4,Energy_4);\n",
+"printf(' \n    _____________________________________________________________________________________');\n",
+"printf(' \n    Total energy required by load for 24hour period (excluding 2hrs at no-load)  =  %d ',W_out_total);\n",
+"printf(' \n    _____________________________________________________________________________________\n\n');\n",
+"\n",
+"printf(' \n d: All-day efficiency = %.1f percent',eta);\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.32: I2_Ic.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-32\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"S_1 = 10 ; // VA rating of small transformer\n",
+"V = 115 ; //  voltage rating of transformer in volt\n",
+"V_2_1 = 6.3 ; // voltage rating of one part of secondary winding in volt\n",
+"V_2_2 = 5.0 ; // voltage rating of other part of secondary winding in volt\n",
+"Z_2_1 = 0.2 ; // impedance of one part of secondary winding in ohm\n",
+"Z_2_2 = 0.15 ; // impedance of other part of secondary winding in ohm\n",
+"\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_2 = V_2_1 + V_2_2 ; // voltage across secondary winding in volt\n",
+"I_2 = S_1 / V_2 ; // Rated secondary current in A when the LV secondaries are\n",
+"// connected in series-aiding\n",
+"\n",
+"// case b\n",
+"I_c = (V_2_1 - V_2_2) / (Z_2_1 + Z_2_2); // Circulating current when LV windings are paralled\n",
+"percent_overload = (I_c / I_2)*100 ; // percent overload produced\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-32 Solution : ');\n",
+"\n",
+"printf(' \n a: Both coils must be series-connected and used to account for the ');\n",
+"printf(' \n    full VA rating of the transformer. Hence,the rated current in 5 V ');\n",
+"printf(' \n    and 6.3 V winding is : \n');\n",
+"printf(' \n    I_2 = %.3f A \n\n', I_2);\n",
+"\n",
+"printf(' \n b: When the windings are paralleled, the net circulating current is ');\n",
+"printf(' \n    the net voltage applied across the total internal impedance of ');\n",
+"printf(' \n    the windings,or :\n');\n",
+"printf(' \n    I_c = %.2f A \n ',I_c);\n",
+"\n",
+"printf(' \n    The percent overload is = %f percent ≃ %.f percent ',percent_overload,percent_overload);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.33: Zeh_Zel_I2rated_I2sc_overload.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-33\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"S = 20 ; // kVA rating of transformer\n",
+"N_1 = 230 ; // Number of primary turns\n",
+"N_2 = 20 ; // Number of secondary turns\n",
+"\n",
+"V_1 = 230 ; // Primary voltage in volt\n",
+"V_2 = 20 ; // Secondary voltage in volt\n",
+"\n",
+"// from Fig.14-31a\n",
+"// HV side SC test data\n",
+"V_sc = 4.5 ; // short circuit voltage in volt\n",
+"I_sc = 87 ; // short circuit current in A\n",
+"P_sc = 250 ; // Power measured in W\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_h = V_sc ;// short circuit voltage in volt on HV side\n",
+"I_h = I_sc ;// short circuit current in A on HV side\n",
+"Z_eh = V_h /I_h ; // Equivalent immpedance reffered to the high side when coils are series connected\n",
+"\n",
+"// case b\n",
+"Z_el = Z_eh * (N_2/N_1)^2 ; //Equivalent immpedance reffered to the low side \n",
+"// when coils are series connected\n",
+"\n",
+"// case c\n",
+"I_2_rated = (S*1000)/V_2 ; // Rated secondary current when coils are series connected\n",
+"\n",
+"// case d\n",
+"I_2_sc = S / Z_el ; // Secondary current when the coils in Fig.14-31a are \n",
+"// short-circuited with rated voltage applied to the HV side \n",
+"\n",
+"percent_overload = (I_2_sc/I_2_rated)*100 ; // percent overload\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-33 Solution : ');\n",
+"\n",
+"printf(' \n    Slight variations in answers are due to non-approximated calculations');\n",
+"printf(' \n    in scilab\n\n');\n",
+"printf(' \n a: Equivalent immpedance reffered to the high side when coils are series connected :');\n",
+"printf(' \n    Z_eh = %f ohm \n ',Z_eh);\n",
+"\n",
+"printf(' \n b: Equivalent immpedance reffered to the low side when coils are series connected :');\n",
+"printf(' \n    Z_el = %f ohm \n ',Z_el);\n",
+"\n",
+"printf(' \n c: Rated secondary current when coils are series connected :');\n",
+"printf(' \n    I_2(rated) = %d A \n',I_2_rated);\n",
+"\n",
+"printf(' \n d: Secondary current when the coils in Fig.14-31a are short-circuited :');\n",
+"printf(' \n    with rated voltage applied to the HV side :');\n",
+"printf(' \n    I_2(sc) = %d A \n',I_2_sc);\n",
+"printf(' \n    The percent overload is = %d percent',percent_overload);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.34: PT_kVA_phase_and_line_currents_kVAtransformers.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-34\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"I_L = 100 ; // Load current in A\n",
+"cos_theta = 0.7 ; // power factor lagging\n",
+"\n",
+"// Y-Δ distribution transformer\n",
+"S = 60 ; // kVA rating of transformer\n",
+"V_1 = 2300 ; // primary voltage in volt\n",
+"V_2 = 230 ; // secondary voltage in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_L = 230 ; // voltage across load in volt\n",
+"P_T = (sqrt(3)*V_L*I_L*cos_theta)/1000 ; // power consumed by the plant in kW\n",
+"kVA_T = P_T/cos_theta ; // apparent power in kVA\n",
+"\n",
+"// case b\n",
+"kVA = S ; // kVA rating of transformer\n",
+"V_p = V_2 ; // phase voltage in volt (delta- connection on  load side)\n",
+"I_P2_rated = (kVA*1000)/(3*V_p); // Rated secondary phase current in A\n",
+"I_L2_rated = sqrt(3)*I_P2_rated ; // Rated secondary line current in A\n",
+"\n",
+"// case c\n",
+"//percent load on each transformer = (load current per line) / (rated current per line)\n",
+"percent_load = I_L / I_L2_rated * 100 ;\n",
+"\n",
+"// case d\n",
+"// subscript d for V_L indicates case d ,V_L\n",
+"V_L_d = 2300 ;\n",
+"I_P1 = (kVA_T*1000)/(sqrt(3)*V_L_d); // primary phase current in A\n",
+"I_L1 = I_P1 ; // primary line current in A(Y-connection)\n",
+"\n",
+"// case e\n",
+"kVA_transformer = kVA / 3 ; // kVA rating of each transformer\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-34 Solution : ');\n",
+"\n",
+"printf(' \n a: power consumed by the plant :\n    P_T = %.1f kW \n ',P_T);\n",
+"printf(' \n    apparent power :\n    kVA_T = %.1f kVA \n',kVA_T);\n",
+"\n",
+"printf(' \n b: Rated secondary phase current :\n    I_P2(rated) = %f A ≃ %.f A \n',I_P2_rated,I_P2_rated);\n",
+"printf(' \n    Rated secondary line current :\n    I_L2(rated) = %f A ≃ %.1f A \n',I_L2_rated,I_L2_rated);\n",
+"\n",
+"printf(' \n c: percent load on each transformer = %.1f percent \n ',percent_load);\n",
+"\n",
+"printf(' \n d: primary phase current :\n    I_P1 = %.f A \n',I_P1);\n",
+"printf(' \n    primary line current :\n    I_L1 = %.f A \n',I_L1);\n",
+"\n",
+"printf(' \n e: kVA rating of each transformer = %d kVA',kVA_transformer);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.35: PT_ST_phase_and_line_currents_kVAtransformers.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-35\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"I_L = 100 ; // Load current in A\n",
+"cos_theta = 0.7 ; // power factor lagging\n",
+"\n",
+"// Δ-Δ distribution transformer\n",
+"S = 60 ; // kVA rating of transformer\n",
+"V_1 = 2300 ; // primary voltage in volt\n",
+"V_2 = 230 ; // secondary voltage in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_L = 230 ; // voltage across load in volt\n",
+"P_T = (sqrt(3)*V_L*I_L*cos_theta)/1000 ; // power consumed by the plant in kW\n",
+"kVA_T = P_T/cos_theta ; // apparent power in kVA\n",
+"\n",
+"// case b\n",
+"kVA = S ; // kVA rating of transformer\n",
+"V_p = V_2 ; // phase voltage in volt \n",
+"I_P2_rated = (kVA*1000)/(3*V_p); // Rated secondary phase current in A\n",
+"I_L2_rated = sqrt(3)*I_P2_rated ; // Rated secondary line current in A\n",
+"\n",
+"// case c\n",
+"//percent load on each transformer = (load current per line) / (rated current per line)\n",
+"percent_load = I_L / I_L2_rated * 100 ;\n",
+"\n",
+"// case d\n",
+"// subscript d for V_L indicates case d ,V_L\n",
+"V_L_d = 2300 ;\n",
+"I_P1 = (kVA_T*1000)/(sqrt(3)*V_L_d); // primary phase current in A\n",
+"I_L1 = sqrt(3)*I_P1 ; // primary line current in A\n",
+"\n",
+"// case e\n",
+"kVA_transformer = kVA / 3 ; // kVA rating of each transformer\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-35 Solution : ');\n",
+"\n",
+"printf(' \n a: power consumed by the plant :\n    P_T = %.1f kW \n ',P_T);\n",
+"printf(' \n    apparent power :\n    kVA_T = %.1f kVA \n',kVA_T);\n",
+"\n",
+"printf(' \n b: Rated secondary phase current :\n    I_P2(rated) = %f A ≃ %.f A \n',I_P2_rated,I_P2_rated);\n",
+"printf(' \n    Rated secondary line current :\n    I_L2(rated) = %f A ≃ %.1f A \n',I_L2_rated,I_L2_rated);\n",
+"\n",
+"printf(' \n c: percent load on each transformer = %.1f percent \n ',percent_load);\n",
+"\n",
+"printf(' \n d: primary phase current :\n    I_P1 = %.f A \n',I_P1);\n",
+"printf(' \n    primary line current :\n    I_L1 = %f A ≃ %.1f A \n',I_L1,I_L1);\n",
+"printf(' \n    The primary line current drawn by a Δ-Δ bank is √3 times the ');\n",
+"printf(' \n    line current drawn by a Y-Δ bank.\n');\n",
+"\n",
+"printf(' \n e: kVA rating of each transformer = %d kVA',kVA_transformer);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.36: find_line_currents_and_their_sum.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-36\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3-phase,3-wire Δ-connected transformer  shown in Fig.14-42\n",
+"V_L = 33 ; // line voltage in kV\n",
+"\n",
+"f = 60 ;// frequency in Hz\n",
+"\n",
+"// power factor\n",
+"PF1 = 1; // unity power factor for I_AB\n",
+"PF2 = 0.7; // 0.7 lagging power factor for I_BC\n",
+"PF3 = 0.9; // 0.9 lagging power factor for I_CA\n",
+"\n",
+"// Calculations\n",
+"V_AB = V_L*exp(%i*(0)*(%pi/180)) ; // line voltage in kV taken as reference voltage\n",
+"\n",
+"V_BC = V_L*exp(%i*(-120)*(%pi/180)) ; // line voltage in kV\n",
+"V_BC_m = abs(V_BC);//V_BC_m=magnitude of V_BC in kV\n",
+"V_BC_a = atan(imag(V_BC) /real(V_BC))*180/%pi - 180 ;//V_BC_a=phase angle of V_BC in degrees\n",
+"// 180 is subtracted from I_BC_a to make it similar to textbook angle\n",
+"\n",
+"V_CA = V_L*exp(%i*(-240)*(%pi/180)) ; // line voltage in kV\n",
+"V_CA_m = abs(V_CA);//V_CA_m=magnitude of V_CA in kV\n",
+"V_CA_a = atan(imag(V_CA) /real(V_CA))*180/%pi - 180 ;//V_CA_a=phase angle of V_CA in degrees\n",
+"// 180 is subtracted from I_BC_a to make it similar to textbook angle\n",
+"\n",
+"theta_1 = acosd(PF1); // PF1 angle\n",
+"theta_2 = acosd(PF2); // PF2 angle\n",
+"theta_3 = acosd(PF3); // PF3 angle\n",
+"\n",
+"\n",
+"I_AB = 10*exp(%i*(theta_1)*(%pi/180)) ; // I_AB current in kA\n",
+"I_AB_m = abs(I_AB);//I_AB_m=magnitude of I_AB in kA\n",
+"I_AB_a = atan(imag(I_AB) /real(I_AB))*180/%pi;//I_AB_a=phase angle of I_AB in degrees\n",
+"\n",
+"I_BC = 15*exp(%i*(-120 - theta_2)*(%pi/180)) ; // I_BC current in kA\n",
+"I_BC_m = abs(I_BC);//I_BC_m=magnitude of I_BC in kA\n",
+"I_BC_a = atan(imag(I_BC) /real(I_BC))*180/%pi - 180;//I_BC_a=phase angle of I_BC in degrees\n",
+"// 180 is subtracted from I_BC_a to make it similar to textbook angle\n",
+"\n",
+"I_CA = 12*exp(%i*(-240 + theta_3)*(%pi/180)) ; // I_CA current in kA\n",
+"I_CA_m = abs(I_CA);//I_CA_m=magnitude of I_CA in kA\n",
+"I_CA_a = 180 + atan(imag(I_CA) /real(I_CA))*180/%pi;//I_CA_a=phase angle of I_CA in degrees\n",
+"// 180 is added to I_BC_a to make it similar to textbook angle\n",
+"\n",
+"// case a\n",
+"I_AC = -I_CA ;\n",
+"I_A = I_AB + I_AC ; // phase current in kA\n",
+"I_A_m = abs(I_A);//I_A_m=magnitude of I_A in kA\n",
+"I_A_a = atan(imag(I_A) /real(I_A))*180/%pi;//I_A_a=phase angle of I_A in degrees\n",
+"\n",
+"// case b\n",
+"I_BA = -I_AB ;\n",
+"I_B = I_BC + I_BA ; // phase current in kA\n",
+"I_B_m = abs(I_B);//I_B_m=magnitude of I_B in kA\n",
+"I_B_a = atan(imag(I_B) /real(I_B))*180/%pi;//I_B_a=phase angle of I_B in degrees\n",
+"\n",
+"// case c\n",
+"I_CB = -I_BC ;\n",
+"I_C = I_CA + I_CB ; // phase current in kA\n",
+"I_C_m = abs(I_C);//I_C_m=magnitude of I_C in kA\n",
+"I_C_a = atan(imag(I_C) /real(I_C))*180/%pi;//I_C_a=phase angle of I_C in degrees\n",
+"\n",
+"// case d\n",
+"phasor_sum = I_A + I_B + I_C ;\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-36 Solution : ');\n",
+"\n",
+"printf(' \n We must first write each of the phase currents in polar form. ');\n",
+"printf(' \n Since reference voltage,V_AB is assumed as 33 <0 kV, we may write\n');\n",
+"\n",
+"printf(' \n I_AB = %d <%d kA (unity PF),\n',I_AB_m,I_AB_a);\n",
+"printf(' \n But I_BC lags V_BC, which is %.f <%d kV',V_BC_m,V_BC_a);\n",
+"printf(' \n by θ = acosd(%.1f) = -%.2f lag, and consequently',PF2,theta_2);\n",
+"printf(' \n I_BC = %.f <%.2f kA \n',I_BC_m,I_BC_a);\n",
+"\n",
+"printf(' \n Similarly,I_CA leads V_CA = %.f <%.f kV',V_CA_m,V_CA_a);\n",
+"printf(' \n by θ = acosd(%.1f) = %.2f lead, and consequently',PF3,theta_3);\n",
+"printf(' \n I_CA = %d <%.2f kA \n',I_CA_m,I_CA_a);\n",
+"\n",
+"printf(' \n Writing three phase currents in comples form yields.\n');\n",
+"printf(' \n I_AB in kA = ');disp(I_AB);\n",
+"printf(' \n I_BC in kA = ');disp(I_BC);\n",
+"printf(' \n I_CA in kA = ');disp(I_CA);\n",
+"\n",
+"printf(' \n From conventional three phase theory for unbalanced Δ-connected loads');\n",
+"printf(' \n and from Fig.14-42, we have\n');\n",
+"\n",
+"printf(' \n a: I_A in kA = ');disp(I_A);\n",
+"printf(' \n    I_A = %.2f <%.2f kA \n',I_A_m,I_A_a);\n",
+"\n",
+"printf(' \n b: I_B in kA = ');disp(I_B);\n",
+"printf(' \n    I_B = %.2f <%.2f kA \n',I_B_m,I_B_a);\n",
+"\n",
+"printf(' \n c: I_C in kA = ');disp(I_C);\n",
+"printf(' \n    I_C = %.2f <%.2f kA \n',I_C_m,I_C_a);\n",
+"\n",
+"printf(' \n d: Phasor sum of the line currents :');\n",
+"printf(' \n    ΣI_L in kA = ');disp(phasor_sum);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.37: kVAcarry_loadtransformer_VVkVA_ratiokVA_increaseload.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-37\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Δ-Δ transformers in Ex.35\n",
+"kVA_1 = 20 ; // kVA rating of transformer 1\n",
+"kVA_2 = 20 ; // kVA rating of transformer 2\n",
+"kVA_3 = 20 ; // kVA rating of transformer 3\n",
+"\n",
+"V_1 = 2300 ; // Primary voltage in volt\n",
+"V_2 = 230 ; // Secondary voltage in volt\n",
+"\n",
+"kVA = 40 ; // kVA supplied by the bank\n",
+"PF = 0.7 ; // lagging power factor at which bank supplies kVA\n",
+"\n",
+"// one defective transformer is removed\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"kVA_transformer = kVA / sqrt(3); // kVA load carried by each transformer\n",
+"\n",
+"// case b\n",
+"percent_ratedload_Tr = kVA_transformer / kVA_1 * 100 ; // percent load carried by each transformer\n",
+"\n",
+"// case c\n",
+"kVA_V_V = sqrt(3)*kVA_1 ; // Total kVA rating of the transformer bank in V-V\n",
+"\n",
+"// case d\n",
+"ratio_banks = kVA_V_V / (kVA_1 + kVA_2 + kVA_3) * 100; // ratio of V-V bank to Δ-Δ bank Tr ratings\n",
+"\n",
+"// case e\n",
+"kVA_Tr = kVA / 3 ;\n",
+"percent_increase_load = kVA_transformer / kVA_Tr * 100 ; // percent increase in load on each transformer when one Tr is removed\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-37 Solution : ');\n",
+"\n",
+"printf(' \n a: kVA load carried by each transformer = %.1f kVA/transformer\n',kVA_transformer);\n",
+"\n",
+"printf(' \n b: percent rated load carried by each transformer = %.1f percent \n',percent_ratedload_Tr);\n",
+"\n",
+"printf(' \n c: Total kVA rating of the transformer bank in V-V = %.2f kVA \n',kVA_V_V);\n",
+"\n",
+"printf(' \n d: ratio of V-V bank to Δ-Δ bank Tr ratings = %.1f percent \n',ratio_banks);\n",
+"\n",
+"printf(' \n e: kVA load carried by each transformer(V-V) = %.2f kVA/transformer\n',kVA_Tr);\n",
+"printf(' \n    percent increase in load on each transformer when one Tr is removed :');\n",
+"printf(' \n     = %.1f percent',percent_increase_load);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.38: IL_alpha_Ia_kVA.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-38\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3-phase SCIM\n",
+"V = 440 ; // rated voltage in volt of SCIM\n",
+"hp = 100 ; // rated power in hp of SCIM\n",
+"PF = 0.8 ; // power factor\n",
+"V_1 = 155 ; // primary voltage in volt of Tr\n",
+"V_2 = 110 ; // secondary voltage in volt of Tr\n",
+"\n",
+"V_a = 110 ; // armature voltage in volt\n",
+"V_L = 440 ; // Load voltage in volt\n",
+"eta = .98 ; // efficiency of the Tr.\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// referring to appendix A-3,Table 430-150 footnotes\n",
+"I_L = 124*1.25 ; // Motor line current in A\n",
+"\n",
+"// case b\n",
+"alpha = V_a/V_L ; // Transformation ratio\n",
+"\n",
+"// case c\n",
+"I_a = (sqrt(3)/2)*( I_L / (alpha*eta) ); // Current in the primary of the scott transformers\n",
+"\n",
+"// case d\n",
+"kVA = (V_a*I_a)/((sqrt(3)/2)*1000); // kVA rating of the main and teaser transformers\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-38 Solution : ');\n",
+"\n",
+"printf(' \n a: Motor line current :\n    I_L = %d A \n ',I_L);\n",
+"\n",
+"printf(' \n b: Transformation ratio :\n    alpha = N_1/N_2 = V_a/V_L = %.2f \n',alpha);\n",
+"\n",
+"printf(' \n c: Current in the primary of the scott transformers :\n    I_a = %.f A \n',I_a);\n",
+"\n",
+"printf(' \n d: kVA rating of the main and teaser transformers :\n    kVA = %.1f kVA',kVA);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.39: VL_ST_Idc_Sac_Sdc_per_line.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-39\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"I_L = 1 ; // Load current in kA\n",
+"V_m = 750 ; // Peak voltage in kV\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_L = (V_m)/sqrt(2); // Max. allowable Vrms in kV that may be applied to the lines using ac\n",
+"\n",
+"// case b\n",
+"S_T_ac = sqrt(3)*V_L*I_L ; // Total 3-phase apparent power in MVA\n",
+"\n",
+"// case c\n",
+"I_rms = I_L ; // rms value of load current in kA\n",
+"I_dc =I_rms*sqrt(2); // Max.allowable current in kA that can be delivered by dc transmission\n",
+"\n",
+"// case d\n",
+"V_dc = V_m ; // dc voltage in kV\n",
+"S_T_dc = V_dc*I_dc ; // Total dc apparent power delivered by two lines in MVA\n",
+"\n",
+"// case e\n",
+"S_ac_line = S_T_ac / 3 ; // Power per ac line\n",
+"\n",
+"// case f\n",
+"S_dc_line = S_T_dc / 2 ; // Power per dc line\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-39 Solution : ');\n",
+"\n",
+"printf(' \n :a Max. allowable Vrms in kV that may be applied to the lines using ac :');\n",
+"printf(' \n    V_L = %.1f kV \n ',V_L);\n",
+"\n",
+"printf(' \n :b Total 3-phase apparent power :\n    S_T = %.1f MVA \n',S_T_ac);\n",
+"\n",
+"printf(' \n :c Max.allowable current in kA that can be delivered by dc transmission :');\n",
+"printf(' \n    I_dc = %.3f kA \n ',I_dc);\n",
+"\n",
+"printf(' \n :d Total dc apparent power delivered by two lines :\n    S_T = %.1f MVA\n',S_T_dc);\n",
+"\n",
+"printf(' \n :e Power per ac line :\n    S/ac line = %.1f MVA/line \n',S_ac_line);\n",
+"\n",
+"printf(' \n :f Power per dc line :\n    S/dc line = %.1f MVA/line \n',S_dc_line);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.3: alpha_Z1_I1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"N_1 = 500 ; // Number of primary turns in the audio output transformer\n",
+"N_2 = 25 ; // Number of secondary turns in the audio output transformer\n",
+"Z_L = 8 ; // Speaker impedance in ohm\n",
+"V_1 = 10 ; // Output voltage of the audio output transformer in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"alpha = N_1/N_2 ; // step-down transformation ratio\n",
+"Z_1 = (alpha)^2 * Z_L ; // Impedance reflected to the transformer primary \n",
+"// at the output of Tr in ohm\n",
+"\n",
+"// case b\n",
+"I_1 = V_1 / Z_1 ; // Primary current in A\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-3 Solution : ');\n",
+"\n",
+"printf(' \n a: Transformation ratio :\n    α = %d\n',alpha);\n",
+"printf(' \n    Impedance reflected to the transformer primary at the output of Tr :');\n",
+"printf(' \n    Z_1 = %d ohm \n ',Z_1);\n",
+"\n",
+"printf(' \n b: Matching transformer primary current :\n    I_1 = %f A',I_1);\n",
+"printf(' \n    I_1 = %.3f mA ',1000 * I_1);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.4: Z2prime_Z3prime_Z1_I1_Pt_V2_P2_V3_P3_Pt.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"N_1 = 600 ; // Number of primary turns \n",
+"N_2 = 150 ; // Some number of secondary turns \n",
+"N_3 = 300 ; // Some number of secondary turns \n",
+"Z_2 = 30 ; // Resistive load in ohm across N_2\n",
+"Z_3 = 15 ; // Resistive load in ohm across N_3\n",
+"R_2 = 30 ;\n",
+"R_3 = 15 ;\n",
+"V_p = 16 ; // Primary applied voltage in volt\n",
+"cos_theta = 1 ; // unity PF\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"Z_2_prime = Z_2 * (N_1/N_2)^2 ; // Impedance reflected to the primary by load Z_2 in ohm\n",
+"\n",
+"// case b\n",
+"Z_3_prime = Z_3 * (N_1/N_3)^2 ; // Impedance reflected to the primary by load Z_3 in ohm\n",
+"\n",
+"// case c\n",
+"// Total impedance reflected to the primary in ohm\n",
+"Z_1 = (Z_2_prime * Z_3_prime) / (Z_2_prime + Z_3_prime) ; \n",
+"\n",
+"// case d\n",
+"I_1 = V_p / Z_1 ; // Total current drawn from the supply in A\n",
+"\n",
+"// case e\n",
+"P_t = V_p * I_1 * cos_theta ; // Total power in W drwan from the supply at unity PF\n",
+"\n",
+"// case f\n",
+"V_2 = V_p * (N_2/N_1) ; // Voltage across Z_2 in volt\n",
+"P_2 = (V_2)^2 / R_2 ; // Power dissipated in load Z_2 in W\n",
+"\n",
+"// case g\n",
+"V_3 = V_p * (N_3/N_1) ; // Voltage across Z_3 in volt\n",
+"P_3 = (V_3)^2 / R_3 ; // Power dissipated in load Z_3 in W\n",
+"\n",
+"// case h\n",
+"P_total = P_2 + P_3 ; // Total power dissipated in both loads in W\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-4 Solution : ');\n",
+"\n",
+"printf(' \n a: Impedance reflected to the primary by load Z_2 : ');\n",
+"printf(' \n    Z_2 = %d ohm \n ',Z_2_prime );\n",
+"\n",
+"printf(' \n b: Impedance reflected to the primary by load Z_3 : ');\n",
+"printf(' \n    Z_3 = %d ohm \n ',Z_3_prime );\n",
+"\n",
+"printf(' \n c: Total impedance reflected to the primary : ');\n",
+"printf(' \n    Z_1 = %.1f ohm \n ',Z_1 );\n",
+"\n",
+"printf(' \n d: Total current drawn from the supply : ');\n",
+"printf(' \n    I_1 = %.1f A \n ',I_1 );\n",
+"\n",
+"printf(' \n e: Total power drawn from the supply at unity PF : ');\n",
+"printf(' \n    P_t = %.1f W \n ',P_t );\n",
+"\n",
+"printf(' \n f: Voltage across Z_2 in volt :\n    V_2 = %d V \n ',V_2 );\n",
+"printf(' \n    Power dissipated in load Z_2 :\n    P_2 = %.2f W \n',P_2 );\n",
+"\n",
+"printf(' \n g: Voltage across Z_3 in volt :\n    V_3 = %d V \n ',V_3 );\n",
+"printf(' \n    Power dissipated in load Z_3 :\n    P_3 = %f W \n',P_3 );\n",
+"\n",
+"printf(' \n h: Total power dissipated in both loads :\n    P_t = %.1f W',P_total);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.5: alpha_N2_N1_ZL.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 100 ; // Power rating of the single channel power amplifier in W\n",
+"Z_p = 3200 ; // Output impedance in ohm of the single channel power amplifier\n",
+"N_p = 1500 ; // Number of primary turns in a tapped impedance-matching transformer\n",
+"Z_L1 = 8 ; // Amplifier output in ohm using a tapped impedance-matching transformer\n",
+"Z_L2 = 4 ; // Amplifier output in ohm using a tapped impedance-matching transformer\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"alpha = sqrt(Z_p/Z_L1) ; // Transformation ratio\n",
+"N_2 = N_p / alpha ; // Total number of secondary turns to match 8 ohm speaker\n",
+"\n",
+"// case b\n",
+"// subscript b for alpha indicates case b\n",
+"alpha_b = sqrt(Z_p/Z_L2) ; // Transformation ratio\n",
+"N_1 = N_p / alpha_b ; // Number of primary turns to match 4 ohm speaker\n",
+"\n",
+"// case c\n",
+"turns_difference = N_2 - N_1 ; // Difference in secondary and primary turns\n",
+"// subscript c for alpha indicates case c\n",
+"alpha_c = (1500/22); // Transformation ratio\n",
+"Z_L = Z_p / (alpha_c)^2 ; // Impedance that must be connected between 4 ohm and\n",
+"// 8 ohm terminals to reflect a primary impedance of 3.2 kilo-ohm\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-5 Solution : ');\n",
+"\n",
+"printf(' \n a: Transformation ratio : \n    α = %d \n ',alpha );\n",
+"printf(' \n    Total number of secondary turns to match 8 ohm speaker : ');\n",
+"printf(' \n    N_2 = %d t \n ',N_2 );\n",
+"\n",
+"printf(' \n b: Transformation ratio : \n    α = %.3f \n ',alpha_b );\n",
+"printf(' \n    Number of primary turns to match 4 ohm speaker : ');\n",
+"printf(' \n    N_1 = %d t \n ',N_1 );\n",
+"\n",
+"printf(' \n c: Difference in secondary and primary turns : ');\n",
+"printf(' \n    N_2 - N_1 = %.f t \n ',turns_difference );\n",
+"printf(' \n    Impedance that must be connected between 4 ohm and 8 ohm ');\n",
+"printf(' \n    terminals to reflect a primary impedance of 3.2 kilo-ohm : ');\n",
+"printf(' \n    Z_L = %.2f ohm ',Z_L );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.6: Z_between_terminals_A_B.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 100 ; // Power rating of the single channel power amplifier in W\n",
+"Z_L1 = 8 ; // Amplifier output in ohm using a tapped impedance-matching transformer\n",
+"Z_L2 = 4 ; // Amplifier output in ohm using a tapped impedance-matching transformer\n",
+"P_servo = 10 ; // Power rating of the servo motor in W\n",
+"Z_servo = 0.7 ; // Impedance of the servo motor in ohm\n",
+"\n",
+"// Calculations\n",
+"root_Z_AB = sqrt(8) - sqrt(4);\n",
+"Z_AB = (root_Z_AB)^2;\n",
+"\n",
+"// Display the resulta\n",
+"disp('Example 14-6 Solution : ');\n",
+"\n",
+"printf(' \n Z_p = %dΩ*(N_p/N_1)^2 = %dΩ*(N_p/N_2)^2\n',Z_L2,Z_L1);\n",
+"printf(' \n     = Z_AB * (N_p/(N_2 - N_1)^2 ) \n');\n",
+"printf(' \n Dividing each of the three numerators by N_p^2 and taking the ');\n",
+"printf(' \n square root of each term, we have\n');\n",
+"\n",
+"printf(' \n √(Z_AB)/(N_2 - N_1) = √(4)/N_1 = √(8)/N_2 \n');\n",
+"printf(' \n √(Z_AB)/(N_2 - N_1) = √(4)/N_1 - √(8)/N_2 \n');\n",
+"printf(' \n yielding, √(Z_AB) = √(8) - √(4) = %f \n',root_Z_AB);\n",
+"printf(' \n which Z_AB = (%f)^2 = %.2f Ω \n',root_Z_AB,Z_AB);\n",
+"\n",
+"    "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.7: alpha_V1_V2_I2_I1_PL_Ps_PT_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 10 * exp(%i * 0 * (%pi/180)); // Supply voltage of the source 10<0 V\n",
+"R_s = 1000 ; // Resistance of the source in ohm \n",
+"R_L = 10 ; // Load resistance in ohm\n",
+"Z_L = R_L ; // Load resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"alpha = sqrt(R_s/R_L) ; // Transformation ratio of the matching transformer for MPT\n",
+"\n",
+"// case b\n",
+"V_1 = V / 2 ; // Terminal voltage in volt of the source at MPT \n",
+"\n",
+"// case c\n",
+"V_2 = V_1 / alpha ; // Terminal voltage in volt across the load at MPT\n",
+"\n",
+"// case d\n",
+"I_2 = V_2 / Z_L ; // Secondary load current in A (method 1)\n",
+"I2 = V / (2*alpha*R_L) ; // Secondary load current in A (method 2)\n",
+"\n",
+"// case e\n",
+"I_1 = I_2 / alpha ; // Primary load current drawn from the source in A (method 1)\n",
+"I1 = V / (2*R_s) ; // Primary load current drawn from the source in A (method 2)\n",
+"\n",
+"// case f\n",
+"P_L = (I_2)^2 * R_L ; // Maximum power dissipated in the load in W\n",
+"\n",
+"// case g\n",
+"P_s = (I_1)^2 * R_s ; // Power dissipated internally within the source in W\n",
+"\n",
+"// case h\n",
+"P_T1 = V * I_1 * cosd(0) ; // Total power supplied by the source in W(method 1)\n",
+"\n",
+"P_T2 = P_L + P_s ; // Total power supplied by the source in W(method 2)\n",
+"\n",
+"// case i\n",
+"P_T = P_T1 ;\n",
+"eta = P_L / P_T * 100 ; // Power transfer efficiency in percent\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-7 Solution : ');\n",
+"\n",
+"printf(' \n a: Transformation ratio of the matching transformer for MPT : ');\n",
+"printf(' \n    α = %d \n ',alpha );\n",
+"\n",
+"printf(' \n b: Terminal voltage of the source at MPT :\n    V_1 = %d V \n',V_1);\n",
+"\n",
+"printf(' \n c: Terminal voltage across the load at MPT :\n    V_2 = %.1f V \n',V_2);\n",
+"\n",
+"printf(' \n d: Secondary load current :');\n",
+"printf(' \n    (method 1) :\n    I_2 = %.2f A = %d mA \n ',I_2, 1000*I_2);\n",
+"\n",
+"printf(' \n    (method 2) :\n    I_2 = %.2f A = %d mA \n ',I2, 1000*I2);\n",
+"\n",
+"printf(' \n e: Primary load current drawn from the source : ');\n",
+"printf(' \n    (method 1) :\n    I_1 = %f A = %d mA  \n ',I_1 , 1000*I_1 );\n",
+"printf(' \n    (method 2) :\n    I_1 = %f A = %d mA  \n ',I1 , 1000*I1 );\n",
+"\n",
+"printf(' \n f: Maximum power dissipated in the load : ');\n",
+"printf(' \n    P_L = %f W = %d mW \n',P_L , 1000*P_L );\n",
+"\n",
+"printf(' \n g: Power dissipated internally within the source : ' );\n",
+"printf(' \n    P_s = %f W = %d mW \n',P_s , 1000*P_s );\n",
+"\n",
+"printf(' \n h: Total power supplied by the source : ');\n",
+"printf(' \n    (method 1) :\n    P_T = %f W = %d mW \n ',P_T1, 1000*P_T1);\n",
+"printf(' \n    (method 2) :\n    P_T = %f W = %d mW \n ',P_T2, 1000*P_T2);\n",
+"\n",
+"printf(' \n i: Power transfer efficiency :\n    η = %d percent ',eta );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.8: PL_alpha_maxPL.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// power transformer\n",
+"V = 20 ; // No-load voltage in volt\n",
+"R_s = 18 ; // Internal resistance of the power amplifier in ohm\n",
+"R_L = 8 ; // Load resistance in ohm(Speaker)\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"V_L = ( R_L / (R_L + R_s) )* V; // Load voltage in volt\n",
+"P_L = (V_L)^2 / R_L ; // Power delivered in W to the speaker when connected \n",
+"// directly to the amplifier\n",
+"\n",
+"// case b\n",
+"alpha = sqrt(R_s/R_L); // Turns ratio of the transformer to maximize speaker power\n",
+"\n",
+"// case c\n",
+"V_2 = V / (2*alpha); // Secondary voltage in volt\n",
+"P_L2 = (V_2)^2 / R_L ; // Maximum power delivered in W to the speaker using matching\n",
+"// transformer of part b\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-8 Solution : ');\n",
+"\n",
+"printf(' \n a; Load voltage :\n    V_L = %.2f V across the 8 Ω speaker\n ',V_L);\n",
+"printf(' \n    Power delivered in W to the speaker when connected directly to the amplifier : ');\n",
+"printf(' \n    P_L = %.2f W \n ', P_L );\n",
+"\n",
+"printf(' \n b: Turns ratio of the transformer to maximize speaker power : ');\n",
+"printf(' \n    α = %.1f \n ', alpha );\n",
+"\n",
+"printf(' \n c: Secondary voltage :\n    V_2 = %f V \n ',V_2 );\n",
+"printf(' \n    Maximum power delivered in W to the speaker using matching ');\n",
+"printf(' \n    transformer of part b :\n    P_L = %f W ',P_L2 );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 14.9: Eh_El_Ih_new_kVA.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 14: TRANSFORMERS\n",
+"// Example 14-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 1 ; // kVA rating of the transformer\n",
+"V_1 = 220 ; // Primary voltage in volt\n",
+"V_2 = 110 ; // Secondary voltage in volt\n",
+"f_o = 400 ; // Frequency in Hz(original frequency)\n",
+"f_f = 60 ; // Frequency in Hz for which the transformer is to be used\n",
+"\n",
+"// Calculations\n",
+"alpha = V_1 / V_2 ; // Transformation ratio\n",
+"// case a\n",
+"E_h = V_1 * (f_f / f_o); // Maximum rms voltage in volt applied to HV side\n",
+"E_1 = E_h ; \n",
+"E_l = E_1 / alpha ; // Maximum rms voltage in volt applied to HV side\n",
+"\n",
+"// case b\n",
+"V_h = V_1 ; // High voltage in volt\n",
+"I_h = kVA * 1000 / V_h ;\n",
+"Vh = E_h ;\n",
+"kVA_new = Vh * I_h ;\n",
+"\n",
+"// Display the results\n",
+"disp('Example 14-9 Solution : ');\n",
+"\n",
+"printf(' \n a: To maintain the same permissible flux density in Eqs.(14-15)');\n",
+"printf(' \n    and (14-16),both voltages of the high and low sides must change ');\n",
+"printf(' \n    in the same proportion as the frequency :  ');\n",
+"printf(' \n    E_h = %d V \n    and,\n    E_l = %.1f V\n',E_h , E_l );\n",
+"\n",
+"printf(' \n b: The original current rating of the transformer is unchanged since');\n",
+"printf(' \n    the conductors still have the same current carrying capacity.');\n",
+"printf(' \n    Thus,\n    I_h = %.3f A\n    and the new kVA rating is',I_h );\n",
+"printf(' \n    V_h*I_h = V_1*I_1 = %d VA = %.2f kVA',kVA_new , kVA_new/1000);"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/2-DYNAMO_CONSTRUCTION_AND_WINDINGS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/2-DYNAMO_CONSTRUCTION_AND_WINDINGS.ipynb
new file mode 100644
index 0000000..cf59439
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/2-DYNAMO_CONSTRUCTION_AND_WINDINGS.ipynb
@@ -0,0 +1,437 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 2: DYNAMO CONSTRUCTION AND WINDINGS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.1: calculate_a_for_lap_and_wave_windings.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"m = 3; // Multipicity of the armature\n",
+"P = 14; // No. of poles\n",
+"\n",
+"// Calculations\n",
+"a_lap = m * P; // No. of parallel paths in the armature for a lap winding\n",
+"a_wave = 2 * m; // No. of parallel paths in the armature for a wave winding\n",
+"\n",
+"// Display the result\n",
+"disp('Example 2-1 Solution : ');\n",
+"\n",
+"printf('\n a: a = %d paths ', a_lap);\n",
+"printf('\n b: a = %d paths ', a_wave);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.2: calculate_generated_emf.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 14; // No. of poles\n",
+"phi = 4.2e6; // Flux per pole\n",
+"S = 60; // Generator speed\n",
+"coils = 420; // No. of coils\n",
+"turns_per_coil = 20;\n",
+"conductors_per_turn = 2;\n",
+"a_lap = 42; // No. of parallel paths in the armature for a lap winding\n",
+"a_wave = 6; // No. of parallel paths in the armature for a wave winding\n",
+"\n",
+"// Calculations\n",
+"Z = coils * turns_per_coil * conductors_per_turn; // No. of conductors\n",
+"E_g_lap = (( phi * Z * S * P ) / ( 60 * a_lap )) * 10 ^ -8; // Generated EMF for\n",
+"// lap winding ( Eq 1-5a)\n",
+"E_g_wave = ( phi * Z * S * P ) / ( 60 * a_wave ) * 10 ^ -8; // Generated EMF for\n",
+"// wave winding ( Eq 1-5a)\n",
+"\n",
+"// Display the result\n",
+"disp('Example 2-2 Solution : ');\n",
+"\n",
+"printf('\n a: Eg = %0.1f V ', E_g_lap);\n",
+"printf('\n b: Eg = %0.1f V ', E_g_wave);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.3: calculate_polespan_p_kp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"slots = 72; // No. of slots\n",
+"P = 4; // No. of poles\n",
+"coils_spanned = 14; // 14 slots are spanned while winding the coils\n",
+"\n",
+"// Calculations\n",
+"Pole_span = slots / P; // Pole span\n",
+"p_not = coils_spanned / Pole_span * 180; // Span of the coil in\n",
+"// electrical degrees\n",
+"funcprot(0) ; // Use  to avoid this message 'Warning : redefining function: beta '\n",
+"beta = (180 - p_not);\n",
+"k_p1 = cosd( beta / 2 ); // Pitch factor using eq(2-7)\n",
+"k_p2 = sind( p_not / 2 ); // Pitch factor using eq(2-8)\n",
+"\n",
+"// Display the results\n",
+"disp('Example 2-3 Solution : ')\n",
+"printf(' \n a: Full-pitch coil span = %d slots/pole ', Pole_span );\n",
+"printf(' \n b: p = %d degrees ', p_not );\n",
+"printf(' \n c: kp = %.2f \t\t eq(2-7)', k_p1 );\n",
+"printf(' \n d: kp = %.2f \t\t eq(2-8)', k_p2 );\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.4: calculate_kp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"fractional_pitch = 13 / 16;\n",
+"slot =96; // No. of slots\n",
+"P = 6; // No. of poles\n",
+"\n",
+"// Calculation\n",
+"k_p = sind( ( fractional_pitch * 180 ) / 2 ); // Pitch factor\n",
+"\n",
+"// Display the result\n",
+"disp('Example 2-4 Solution : ')\n",
+"printf('\n kp = %.4f ', k_p );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.5: find_alpha_n_theta.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 12; // No. of poles\n",
+"theta = 360; // No. of mechanical degrees of rotation\n",
+"alpha_b = 180; // No. of electrical degrees for finding case b in the question\n",
+"\n",
+"// Calculations\n",
+"alpha = ( P * theta ) / 2; // No. of electrical degrees in one revolution\n",
+"n = alpha / 360; // No. of ac cycles\n",
+"theta_b = ( 2 * alpha_b ) / P; // No. of mechanical degrees of rotation\n",
+"// for finding case b in the question\n",
+"\n",
+"// Display the results\n",
+"disp('Example 2-5 Solution : ')\n",
+"printf('\n a: alpha = %d degrees', alpha);\n",
+"printf('\n    n = %d cycles ', n);\n",
+"printf('\n b: theta = %d mechanical degrees ', theta_b );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.6: find_n_alpha_kd_for_different_number_of_slots.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 4;// No. of poles\n",
+"phi = 3; // No. of phases\n",
+"slots_(1) = 12; // No. of slots for case 1\n",
+"slots_(2) = 24; // No. of slots for case 2\n",
+"slots_(3) = 48; // No. of slots for case 3\n",
+"slots_(4) = 84; // No. of slots for case 4\n",
+"\n",
+"// Calculations\n",
+"electrical_degrees = 180 * 4;\n",
+"i=1; // where i is case subscript .eg case1, case2, etc\n",
+"\n",
+"  while i<=4  \n",
+"    alpha_(i) = electrical_degrees / slots_(i); // electrical degrees  \n",
+"    // per slots for case i\n",
+"    n_(i) = slots_(i) / ( P * phi ); // No. of ac cycles for case 1\n",
+"    k_d(i) = sind( n_(i)*( alpha_(i) / 2 ) ) / ( n_(i) * sind( alpha_(i) / 2));\n",
+"    i=i+1;\n",
+"  end;\n",
+"     \n",
+"// Display the results\n",
+"disp('Example 2-6 Solution : ')\n",
+"printf('\n a:');\n",
+" i=1; // where i is case subscript .eg case1, case2, etc\n",
+"   \n",
+"   while i<=4\n",
+"        printf('\n \t %d: alpha = %.2f degrees/slot', i , alpha_(i) );\n",
+"        printf('\n\t    n = %d slots/pole-phase ', n_(i) );\n",
+"        printf('\n\t    kd = %.3f ', k_d(i));\n",
+"        printf('\n');\n",
+"        i=i+1; \n",
+"   end;\n",
+"   \n",
+"printf('\n\n\n b: ');\n",
+"printf('\n \t \t n \t alpha in degrees  \t\t kd ');\n",
+"printf('\n \t __________________________________________________________' );\n",
+"i=1;\n",
+"\n",
+" while i<=4\n",
+"    printf('\n \t \t %d \t %.2f \t\t\t\t%.3f ', n_(i) , alpha_(i) , k_d(i) );\n",
+"    i = i +1;\n",
+"  end;\n",
+"    printf('\n \t __________________________________________________________' );\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.7: calculate_Eg_Np_kd_kp_Egp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"slots = 72; // No. of slots\n",
+"P = 6; // No. of poles\n",
+"phase =3; // three phase stator armature\n",
+"N_c = 20; // Number of turns per coil\n",
+"pitch = 5 / 6;\n",
+"phi = 4.8e+6; // flux per pole in lines\n",
+"S = 1200; // Rotor speed\n",
+"\n",
+"// Calculations\n",
+"f = ( P * S )/ 120; // Frequency of rotor\n",
+"\n",
+"E_g_percoil = 4.44 * phi * N_c * f *10 ^ -8; // Generated effective voltage\n",
+"// per coil of a full pitch coil\n",
+"\n",
+"N_p = ( slots / phase ) * N_c; // Total number of turns per phase\n",
+"\n",
+"n = slots / ( phase * P ); // No. os slots per pole per phase\n",
+"\n",
+"alpha = ( P * 180 ) / slots; // No. of electrical degrees between adjacent slots\n",
+"\n",
+"k_d = sind( n * alpha / 2 ) / ( n * sind( alpha / 2 ) ); // Distribution factor\n",
+"\n",
+"span = pitch * 180; // Span of the coil in electrical degrees\n",
+"\n",
+"k_p = sind( span / 2 ); // Pitch factor\n",
+"\n",
+"E_gp = 4.44 * phi * N_p * f * k_p * k_d * 10 ^ -8; // Total generated voltage\n",
+"// per phase considering kp and kd\n",
+"\n",
+"// Display the result\n",
+"disp('Example 2-7 Solution : ')\n",
+"printf('\n a: Eg/coil = %.2f V/coil', E_g_percoil );\n",
+"printf('\n b: Np = %d turns/phase ', N_p );\n",
+"printf('\n c: kd = %.3f ', k_d );\n",
+"printf('\n d: kp = %.3f ', k_p );\n",
+"printf('\n e: Egp = %.2f V/phase ', E_gp );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 2.8: calculate_f_S_omega.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 2: Dynamo Construction and Windings\n",
+"// Example 2-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 8; // No. of poles\n",
+"S = 900; // Speed in revolutions / minute\n",
+"f_1 = 50; // Frequency of generated voltage from generator 1\n",
+"f_2 = 25; // Frequency of generated voltage from generator 2\n",
+"\n",
+"// Calculations\n",
+"f = ( P * S ) / 120; // Frequency of the generated voltage\n",
+"S_1 = ( 120 * f_1 ) / P; // Speed of generator(rpm) 1 to generate 50 Hz voltage\n",
+"S_2 = ( 120 * f_2 ) / P; // Speed of generator(rpm) 2 to generate 25 Hz voltage\n",
+"omega_1 = ( 4 * %pi * f_1 ) / P; // Speed of generator 1 in rad/s\n",
+"omega_2 = ( 4 * %pi * f_2 ) / P; // Speed of generator 2 in rad/s\n",
+"\n",
+"// Display the result\n",
+"disp('Example 2-8 Solution : ')\n",
+"printf('\n a: f = %d Hz ', f );\n",
+"printf('\n b: S1 = %d rpm \n    S2 = %d rpm ', S_1, S_2 );\n",
+"printf('\n c: omega1 = %f rad/s \n    omega2 = %f rad/s', omega_1, omega_2 );"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/3-DC_DYNAMO_VOLTAGE_RELATIONS_DC_GENERATORS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/3-DC_DYNAMO_VOLTAGE_RELATIONS_DC_GENERATORS.ipynb
new file mode 100644
index 0000000..77c363f
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/3-DC_DYNAMO_VOLTAGE_RELATIONS_DC_GENERATORS.ipynb
@@ -0,0 +1,501 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 3: DC DYNAMO VOLTAGE RELATIONS DC GENERATORS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.10: determine_approx_size_of_dynamo.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kW= 50; // Power rating of the dynamo\n",
+"V = 125; // Rated voltage in V\n",
+"S = 1800; // Speed of the dynamo in rpm\n",
+"I_f =20; // Exciting field current\n",
+"Max_temp_rise = 25; // Maximum Temperature rise in degree celsius\n",
+"I_l = 400; // Load Current in A\n",
+"// INSULATION CLASS A\n",
+"// COMPOUND WINDING \n",
+"\n",
+"// Display the result\n",
+"disp('Example 3-10 Solution : ')\n",
+"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. ');\n",
+"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.');\n",
+"printf('\n\n c: Half the size. ');\n",
+"printf('\n\n d: Same size. ');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.1: calculate_I1_If_Ia_Eg.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kW = 150; // Power rating of Shunt generator in kW\n",
+"V_1 = 250; // Voltage rating of Shunt generator in V\n",
+"V_a = V_1; // Voltage rating of Shunt generator in V\n",
+"R_f = 50; // Field resistance in ohm\n",
+"R_a = 0.05; // Armature resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"I_1 = ( kW * 1000 ) / V_1; // Full-load line current flowing to the load in A\n",
+"I_f = V_1 / R_f; // Field current in A\n",
+"I_a = I_f + I_1; // Armature current in A\n",
+"E_g = V_a + I_a * R_a; // Full load generated voltage in V\n",
+"\n",
+"// Display the results\n",
+"disp('Example 3-1 Solution : ')\n",
+"printf('\n a: I1 = %d A ', I_1 );\n",
+"printf('\n b: If = %d A ', I_f );\n",
+"printf('\n c: Ia = %d A ', I_a );\n",
+"printf('\n d: Eg = %.2f A ', E_g );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.2: calculate_Rd_Eg.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kW =100; // Power rating of the generator in kW\n",
+"V_1 = 500; // Voltage rating of hte generator in V\n",
+"R_a = 0.03; // Armature resistance in ohm\n",
+"R_f = 125; // Shunt field resistance in ohm\n",
+"R_s = 0.01; // Series field resistance in ohm\n",
+"I_d = 54; // Diverter current in A\n",
+"\n",
+"// Calculations\n",
+"I_1 = ( kW * 1000 ) / V_1; // Full-load line current flowing to the load in A\n",
+"I_f = V_1 / R_f; // Shunt Field current in A\n",
+"I_a = I_f + I_1; // Armature current in A\n",
+"I_s = I_a - I_d; // Series Field current in A\n",
+"R_d = ( I_s * R_s ) / I_d; // Diverter resistance in ohm\n",
+"E_g = V_1 + I_a * R_a + I_s * R_s; // Generated voltage at full load in V\n",
+"\n",
+"// Display the results\n",
+"disp('Example 3-2 Solution : ')\n",
+"printf('\n a: Rd = %.4f ohm ', R_d );\n",
+"printf('\n b: Eg = %.2f V ', E_g );\n",
+"\n",
+"\n",
+"\n",
+"\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.3: calculate_Vnoload.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"E_orig = 150; // Armature voltage of the generator  in V\n",
+"S_orig = 1800; // Speed of the generator in rpm \n",
+"S_final_a =2000; // Increased Speed of the generator in rpm for case a\n",
+"S_final_b =1600; // Increased Speed of the generator in rpm for case b\n",
+"\n",
+"// Calculations\n",
+"E_final_a = E_orig * ( S_final_a / S_orig ); // No- load voltage of the generator \n",
+"// generator  in V for case a\n",
+"E_final_b = E_orig * ( S_final_b / S_orig ); // No- load voltage of the generator \n",
+"// generator  in V for case b\n",
+"\n",
+"// Display the results\n",
+"disp('Example 3-3 Solution : ')\n",
+"printf('\n a: Efinal = %.1f V ', E_final_a );\n",
+"printf('\n b: Efinal = %.1f V ', E_final_b );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.4: calculate_E.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"S_final = 1200; // Speed of th generator in rpm\n",
+"E_orig_a = 64.3; // Armature voltage of the generator  in V for case a\n",
+"E_orig_b = 82.9; // Armature voltage of the generator  in V for case b\n",
+"E_orig_c = 162.3; // Armature voltage of the generator  in V for case c\n",
+"\n",
+"S_orig_a = 1205; // Varied Speed of the generator in rpm for case a\n",
+"S_orig_b = 1194; // Varied Speed of the generator in rpm for case b\n",
+"S_orig_c = 1202; // Varied Speed of the generator in rpm for case c\n",
+"\n",
+"// Calculations\n",
+"E_1 = E_orig_a * ( S_final / S_orig_a ); // No- load voltage of the generator \n",
+"// generator  in V for case a\n",
+"E_2 = E_orig_b * ( S_final / S_orig_b ); // No- load voltage of the generator \n",
+"// generator  in V for case b\n",
+"E_3 = E_orig_c * ( S_final / S_orig_c ); // No- load voltage of the generator \n",
+"// generator  in V for case c\n",
+"\n",
+"// Display the results\n",
+"disp('Example 3-4 Solution : ')\n",
+"printf('\n a: E1 = %.1f V at %d rpm ', E_1, S_final );\n",
+"printf('\n b: E2 = %.1f V at %d rpm ', E_2, S_final );\n",
+"printf('\n c: E3 = %.1f V at %d rpm ', E_3, S_final );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.5: calculate_Ia_Eg.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V = 125; // Rated voltage of the shunt generator in V\n",
+"R_a = 0.15; // Armature resistance in ohm\n",
+"V_a = 0; // Shunt generator is loaded progressively until the terminal voltage\n",
+"// across the load is zero volt\n",
+"I_1 = 96; // Load current in A\n",
+"I_f = 4; // Field current in A\n",
+"\n",
+"// Calculations\n",
+"I_a = I_f + I_1; // Armature current in A\n",
+"E_g = V_a + I_a * R_a ; // Voltage generated in the armature in V\n",
+"\n",
+"// Display the results\n",
+"disp('Example 3-5 Solution : ')\n",
+"printf('\n Ia = %d A ', I_a );\n",
+"printf('\n Eg = %d V ', E_g );\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.6: calculate_VR.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_n1 = 135; // No load voltage of the shunt generator in V\n",
+"V_f1 = 125; // Full load voltage of the shunt generator in V\n",
+"\n",
+"// Calculation\n",
+"VR = ( V_n1 - V_f1 ) / V_f1 * 100; // Percentage voltage regulation\n",
+"\n",
+"// Display the result\n",
+"disp('Example 3-6 Solution : ')\n",
+"printf(' \n VR = %d percent ', VR );\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.7: calculate_Vnoload.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"VR = 0.105; // voltage regulation\n",
+"V_f1 = 250; // Full load voltage of the shunt generator in V\n",
+"\n",
+"// Calculation\n",
+"V_n1 = V_f1 + ( V_f1 * VR ); // No-load voltage of the generator in V\n",
+"\n",
+"// Display the result\n",
+"disp('Example 3-7 Solution : ')\n",
+"printf('\n Vn1 = %.1f V ', V_n1 );\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.8: calculate_IsNs_Rd.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"N_f = 1000; // Shunt field winding turns\n",
+"N_s = 4; // Series field winding turns\n",
+"I_f = 0.2; // Field current in A\n",
+"I_a = 80; // Full load armature current in A\n",
+"R_s =0.05; // Series field resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"deba_I_f_N_f = I_f * N_f; \n",
+"I_s_N_s = deba_I_f_N_f; // Series field ampere-turns\n",
+"I_s =( I_s_N_s ) / N_s; // Desired current in A in the series field required to\n",
+"// produce voltage rise\n",
+"I_d = I_a - I_s; // Diverter current in A\n",
+"R_d = ( I_s * R_s ) / I_d; // Diverter resistance in ohm\n",
+"\n",
+"// Display the result\n",
+"disp('Example 3-8 Solution : ')\n",
+"printf('\n a: IsNs = %d At ', I_s_N_s );\n",
+"printf('\n b: Rd = %.4f ohm ', R_d );\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 3.9: calculate_Rd_Vnl_Vfl.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 3: DC Dynamo Voltage Relations - DC Generators\n",
+"// Example 3-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kW = 60; // Power rating of the generator in kW\n",
+"V = 240; //Voltage rating of the generator in V\n",
+"I_f = 3; // Increase in the field current in A\n",
+"OC_V = 275; // Over Compounded Voltage in V\n",
+"I_l = 250; // Rated load current in A\n",
+"N_f = 200; // No. of turns per pole in the shunt field winding \n",
+"N_s = 5; // No. of turns per pole in the series field winding \n",
+"R_f = 240; //  Shunt field resistance in ohm\n",
+"R_s = 0.005; //  Series field resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"deba_I_f_N_f = I_f * N_f; \n",
+"I_s_N_s = deba_I_f_N_f; // Series field ampere-turns\n",
+"I_s =( I_s_N_s ) / N_s; // Desired current in A in the series field required to\n",
+"// produce voltage rise\n",
+"I_d = I_l - I_s; // Diverter current in A\n",
+"R_d = ( I_s * R_s ) / I_d; // Diverter resistance in ohm\n",
+"NL_MMF = ( V / R_f )* N_f; // No-load MMF\n",
+"I_f_N_f = NL_MMF;\n",
+"FL_MMF = I_f_N_f + I_s_N_s; // Full-load MMF\n",
+"\n",
+"// Display the result\n",
+"disp('Example 3-9 Solution : ')\n",
+"printf('\n a: Rd = %.5f ohm ', R_d );\n",
+"printf('\n b: No-load MMF = %d At/pole ', NL_MMF );\n",
+"printf('\n    Full-load MMF = %d At/pole ', FL_MMF );"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/4-DC_DYNAMO_TORQUE_RELATIONS_DC_MOTORS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/4-DC_DYNAMO_TORQUE_RELATIONS_DC_MOTORS.ipynb
new file mode 100644
index 0000000..65650ec
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/4-DC_DYNAMO_TORQUE_RELATIONS_DC_MOTORS.ipynb
@@ -0,0 +1,1247 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 4: DC DYNAMO TORQUE RELATIONS DC MOTORS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.10: calculate_power_developed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"I_a_1 = 38 ; // Armature current in A @ full-load as per example 4-8a\n",
+"E_c_1 = 109.4 ; // Back EMF in volt @ full-load as per example 4-8a\n",
+"S_1 = 1800 ; // Speed in rpm @ full-load as per example 4-8a\n",
+"\n",
+"I_a_2 = 19 ; // Armature current in A @ half-loadas per example 4-8a\n",
+"E_c_2 = 113.2 ; // Back EMF in volt @ half-load as per example 4-8a\n",
+"S_2 = 1863 ; // Speed in rpm @ half-load as per example 4-8a\n",
+"\n",
+"I_a_3 = 47.5 ; // Armature current in A @ 125% overload as per example 4-8b\n",
+"E_c_3 = 107.5 ; // Back EMF in volt @ 125% overload as per example 4-8b\n",
+"S_3 = 1769 ; // Speed in rpm @ 125% overload as per example 4-8b\n",
+"\n",
+"I_a_4 = 63.6 ; // Armature current in A @ overload as per example 4-9\n",
+"E_c_4 = 104.3 ; // Back EMF in volt @ overload as per example 4-9\n",
+"S_4 = 1532 ; // Speed in rpm @ overload as per example 4-9\n",
+"\n",
+"// Calculations\n",
+"P_d_1 = E_c_1 * I_a_1 ; // Armature power developed @ full-load\n",
+"\n",
+"P_d_2 = E_c_2 * I_a_2 ; // Armature power developed @ half-load\n",
+"\n",
+"P_d_3 = E_c_3 * I_a_3 ; // Armature power developed @ 125% overload\n",
+"\n",
+"P_d_4 = E_c_4 * I_a_4 ; // Armature power developed @ overload\n",
+"\n",
+"// Display the results\n",
+"disp(' Example 4-10 Solution : ');\n",
+"printf('\n Example \t Ia \t Ec \t Speed \t Pd or (Ec*Ia)');\n",
+"printf('\n _______________________________________________________________________');\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);\n",
+"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);\n",
+"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);\n",
+"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);\n",
+"printf('\n _______________________________________________________________________');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.11: convert_torque_readings_into_Nm_and_lbft.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"T_a = 6.5; // Torque in dyne-centimeters\n",
+"T_b = 10.6; // Torque in gram-centimeters\n",
+"T_c = 12.2; // Torque in ounce-inches\n",
+"\n",
+"// Calculations\n",
+"T_a_Nm = T_a * 1.416e-5 * 7.0612e-3 ; // Torque T_a in N-m \n",
+"T_b_Nm = T_b * ( 1 / 72.01 ) * 7.0612e-3 ; // Torque T_b in N-m \n",
+"T_c_Nm = T_c * 7.0612e-3 ; // Torque T_c in N-m \n",
+"\n",
+"T_a_lbft = T_a * 1.416e-5 * 5.208e-3; // Torque T_a in lb-ft\n",
+"T_b_lbft = T_b * ( 1 / 72.01 ) * 5.208e-3; // Torque T_b in lb-ft\n",
+"T_c_lbft = T_c * 5.208e-3; // Torque T_c in lb-ft\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-11 Solution : ');\n",
+"printf(' \n a : T = %.1e N-m ', T_a_Nm );\n",
+"printf(' \n     T = %.1e lb-ft \n ', T_a_lbft );\n",
+"\n",
+"printf(' \n b : T = %.2e N-m ', T_b_Nm );\n",
+"printf(' \n     T = %.1e lb-ft \n ', T_b_lbft );\n",
+"\n",
+"printf(' \n c : T = %.3e N-m ', T_c_Nm );\n",
+"printf(' \n     T = %.2e lb-ft \n ', T_c_lbft );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.12: calculate_Ist_and_percentage_of_load_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-12\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_a = 120 ; // Rated terminal voltage of dc shunt notor in volt\n",
+"R_a = 0.2 ; // Armature resistance in ohm\n",
+"BD = 2 ; // Brush drop in volt\n",
+"I_a = 75 ; // Full load armature current in A\n",
+"\n",
+"// Calculations\n",
+"I_st = ( V_a - BD ) / R_a ; // Current @ the instant of starting in A\n",
+"percentage = I_st / I_a * 100 ; // Percentage at full load\n",
+"\n",
+"// Display the results \n",
+"disp(' Example 4-12 Solution : ');\n",
+"printf(' \n Ist = %d A ( Back EMF is zero )',I_st );\n",
+"printf(' \n Percentage at full load = %d percent ', percentage );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.13: calculate_Rs_at_various_back_Emfs_and_Ec_at_zero_Rs.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-13\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_a = 120 ; // Rated terminal voltage of dc shunt notor in volt\n",
+"R_a = 0.2 ; // Armature resistance in ohm\n",
+"BD = 2 ; // Brush drop in volt\n",
+"I_a = 75 ; // Full load armature current in A\n",
+"I_a_new = 1.5 * I_a ; // armature current in A at 150% rated load \n",
+"\n",
+"E_c_a = 0 ; // Back EMF at starting\n",
+"E_c_b = ( 25 / 100 ) * V_a ; // Back EMF in volt is 25% of Va @ 150% rated load\n",
+"E_c_c = ( 50 / 100 ) * V_a ; // Back EMF in volt is 50% of Va @ 150% rated load\n",
+"\n",
+"// Calculations\n",
+"R_s_a = ( V_a - E_c_a - BD ) / I_a_new - R_a ; // Ra tapping value at starting \n",
+"// in ohm\n",
+"R_s_b = ( V_a - E_c_b - BD ) / I_a_new - R_a ; // Ra tapping value @ 25% of Va \n",
+"//  in ohm\n",
+"R_s_c = ( V_a - E_c_c - BD ) / I_a_new - R_a ; // Ra tapping value @ 50% of Va \n",
+"// in ohm\n",
+"E_c_d = V_a - ( I_a * R_a + BD ) ; // Back EMF @ full-load without starting resistance\n",
+"\n",
+"// Display the results\n",
+"disp(' Example 4-13 Solution : ');\n",
+"printf(' \n a: At starting, Ec is zero ');\n",
+"printf(' \n    Rs = %.2f ohm \n ', R_s_a );\n",
+"\n",
+"printf(' \n b: When back EMF in volt is 25 percent of Va @ 150 percent rated load ');\n",
+"printf(' \n    Rs = %.3f ohm \n ', R_s_b );\n",
+"\n",
+"printf(' \n c: When back EMF in volt is 50 percent of Va @ 150 percent rated load ');\n",
+"printf(' \n    Rs = %.3f ohm \n ', R_s_c );\n",
+"\n",
+"printf(' \n d: Back EMF at full-load without starting resistance ');\n",
+"printf(' \n    Ec = %d V ', E_c_d );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.14: calculate_field_flux_in_percent_and_final_torque_developed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-14\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Cumulative DC compound motor acting as shunt motor\n",
+"T_orig = 160 ; // Original torque developed in lb.ft\n",
+"I_a_orig = 140 ; // Original armature current in A\n",
+"phi_f_orig = 1.6e+6 ; // Original field flux in lines\n",
+"\n",
+"// Reconnected as a cumulative DC compound motor\n",
+"T_final_a = 190 ; // Final torque developed in lb.ft (case a)\n",
+"\n",
+"// Calculations\n",
+"phi_f = phi_f_orig * ( T_final_a / T_orig ) ; // Field flux in lines\n",
+"percentage = ( phi_f / phi_f_orig ) * 100 - 100 ; // percentage increase in flux\n",
+"\n",
+"phi_f_final = 1.1 * phi_f ; // 10% increase in load causes 10% increase in flux\n",
+"I_a_b = 154 ; // Final armature current in A (case b)\n",
+"T_f = T_final_a * ( I_a_b / I_a_orig ) * ( phi_f_final / phi_f ) ; \n",
+"// Final torque developed\n",
+"\n",
+"// Display the results\n",
+"disp(' Example 4-14 Solution : ');\n",
+"printf(' \n a: phi_f = %.1e lines \n ', phi_f );\n",
+"printf(' \n    Percentage of flux increase = %.1f percent \n ', percentage );\n",
+"\n",
+"printf(' \n b: The final field flux is 1.1 * 1.9 * 10 ^ 6 lines ' );\n",
+"printf(' \n    (due to the 10 percent increase in load).The final torque is\n');\n",
+"printf(' \n    T_f = %.1f lb-ft ', T_f );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.15: calculate_torque_developed_for_varying_flux_and_Ia.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-15\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"I_a_orig = 25 ; // Original armature current in A\n",
+"I_a_final = 30 ; // Final armature current in A\n",
+"T_orig = 90 ; // Original torque developed in lb.ft\n",
+"phi_orig = 1.0 ; // Original flux\n",
+"phi_final = 1.1 ; // Final flux\n",
+"\n",
+"// Calculations\n",
+"T_a = T_orig * ( I_a_final / I_a_orig ) ^ 2 ; // Final torque developed if field\n",
+"// is unsaturated\n",
+"T_b = T_orig * ( I_a_final / I_a_orig ) * ( phi_final / phi_orig ) ; \n",
+"// Final torque developed when Ia rises to 30 A and flux by 10%\n",
+"\n",
+"// Display the results\n",
+"disp(' Example 4-15 Solution : ' );\n",
+"printf(' \n a: T = %.1f lb-ft \n ', T_a );\n",
+"printf(' \n b: T = %.1f lb-ft ', T_b );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.16: calculate_speed_at_rated_load_and_P_and_hp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-16\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_a = 230 ; // Rated armature voltage in volt\n",
+"P = 10 ; // Rated power in hp\n",
+"S = 1250 ; // Rated speed in rpm\n",
+"R_A = 0.25 ; // Armature resistance in ohm\n",
+"R_p = 0.25 ; // Interpolar resistance\n",
+"BD = 5 ; // Brush voltage drop in volt\n",
+"R_s = 0.15 ; // Series field resistance in ohm\n",
+"R_sh = 230 ; // Shunt field resistance in ohm\n",
+"\n",
+"// shunt connection\n",
+"I_l = 54 ; // Line current in A at rated load \n",
+"I_ol = 4 ; // No-load line current in A\n",
+"S_o = 1810 ; // No-load speed in rpm\n",
+"\n",
+"// Calculations\n",
+"R_a = R_A + R_p ; // Effective armature resistance in ohm\n",
+"I_f = V_a / R_sh ; // Field current in A ( Shunt connection )\n",
+"I_a = I_ol - I_f ; // Armature current in A\n",
+"\n",
+"E_c_o = V_a - ( I_a * R_a + BD ); // No-load BACK EMF in volt\n",
+"E_c_full_load = V_a - ( I_l * R_a + BD ); // No-load BACK EMF in volt at full-load\n",
+"\n",
+"S_r = S_o * ( E_c_full_load / E_c_o ); // Speed at rated load\n",
+"\n",
+"P_d = E_c_full_load * I_l ; // Internal power in watts\n",
+"hp = P_d / 746 ; // Internal horse power\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-16 Solution : ');\n",
+"printf(' \n a: S_r = %d rpm\n ', S_r );\n",
+"printf(' \n b: P_d = %d W ', P_d );\n",
+"printf(' \n    hp = %.1f hp ', hp );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.17: calculate_speed_torque_and_horsepower.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-17\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_a = 230 ; // Rated armature voltage in volt\n",
+"P = 10 ; // Rated power in hp\n",
+"S = 1250 ; // Rated speed in rpm\n",
+"R_A = 0.25 ; // Armature resistance in ohm\n",
+"R_p = 0.25 ; // Interpolar resistance\n",
+"BD = 5 ; // Brush voltage drop in volt\n",
+"R_s = 0.15 ; // Series field resistance in ohm\n",
+"R_sh = 230 ; // Shunt field resistance in ohm\n",
+"phi_1 = 1 ;// Original flux per pole\n",
+"\n",
+"// Long-shunt cumulative connection\n",
+"I_l = 55 ; // Line current in A at rated load \n",
+"phi_2 = 1.25 ; // Flux increased by 25% due to long-shunt cumulative connection\n",
+"I_ol = 4 ; // No-load line current in A\n",
+"S_o = 1810 ; // No-load speed in rpm\n",
+"\n",
+"// Calculations\n",
+"R_a = R_A + R_p ; // Effective armature resistance in ohm\n",
+"I_f = V_a / R_sh ; // Field current in A in shunt winding\n",
+"I_a = I_ol - I_f ; // Armature current in A for shunt connection\n",
+"E_c_o = V_a - ( I_a * R_a + BD ); // No-load BACK EMF in volt for shunt connection\n",
+"E_c_o1 = V_a - ( I_a * R_a + I_a * R_s + BD ); // No-load BACK EMF in volt for\n",
+"// long shunt cumulative connection\n",
+"S_n1 = S_o * ( E_c_o1 / E_c_o ); // Speed at no load\n",
+"\n",
+"I_f = V_a / R_sh ; // Field current in A in shunt winding\n",
+"I_a_lsh = I_l - I_f ; // Armature current in A\n",
+"E_c_full_load = V_a - ( I_a_lsh * R_a + BD ); // No-load BACK EMF in volt at\n",
+"// full-load for long-shunt cumulative connection \n",
+"\n",
+"E_c_full_load_lsh = V_a - ( I_a_lsh * R_a + I_a_lsh * R_s + BD ); // BACK EMF in volt \n",
+"// at full-load for long-shunt cumulative motor \n",
+"\n",
+"S_r = S_o * ( E_c_full_load / E_c_o ); // Speed at rated load for shunt connection\n",
+"S_r_lsh = S_n1 * ( E_c_full_load_lsh / E_c_o1 ) * ( phi_1 / phi_2 ); \n",
+"// Speed at rated load for shunt connection\n",
+"\n",
+"P_d = E_c_full_load * I_a_lsh ; // Internal power in watts\n",
+"hp = P_d / 746 ; // Internal horse power\n",
+"\n",
+"T_shunt = ( hp * 5252 ) / S_r ; // Internal torque @ full-load for shunt motor\n",
+"\n",
+"I_a1 = I_a_lsh; // Armature current for shunt motor in A\n",
+"I_a2 = I_a_lsh; // Armature current for long-shunt cumulative motor in A\n",
+"T_comp = T_shunt * ( phi_2 / phi_1 ) * ( I_a2 / I_a1); // Internal torque \n",
+"// at full-load for long-shunt cumulative motor in A\n",
+"\n",
+"Horsepower = ( E_c_full_load_lsh * I_a_lsh ) / 746 ; // Internal horsepower of\n",
+"// compound motor based on flux increase\n",
+"\n",
+"// Display the results\n",
+"disp(' Example 4-17 Solution : ');\n",
+"printf(' \n a: S_n1 = %d rpm \n', S_n1 );\n",
+"printf(' \n b: S_r = %d rpm \n', S_r_lsh );\n",
+"printf(' \n c: Internal torque of shunt motor at full-load : ');\n",
+"printf(' \n    T_shunt = %.2f lb-ft ', T_shunt );\n",
+"printf(' \n    T_comp = %.2f lb-ft \n', T_comp );\n",
+"printf(' \n d: Horsepower = %.1f hp \n', Horsepower );\n",
+"printf(' \n e: The internal horsepower exceeds the rated horsepower because ');\n",
+"printf(' \n    the power developed in the motor must also overcome the internal');\n",
+"printf(' \n    mechanical rotational losses. ');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.18: calculate_speed_with_and_without_diverter.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-18\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 25 ; // Power rating of a series motor in hp\n",
+"V_a = 250 ; // Rated voltage in volt\n",
+"R_a = 0.1 ; // Armature ckt resistance in ohm\n",
+"BD = 3 ; // Brush voltage drop in volt\n",
+"R_s = 0.05 ; // Series field resistance in ohm\n",
+"I_a1 = 85 ; // Armature current in A (case a)\n",
+"I_a2 = 100 ; // Armature current in A (case b)\n",
+"I_a3 = 40 ; // Armature current in A (case c)\n",
+"S_1 = 600 ; // Speed in rpm at current I_a1\n",
+"R_d = 0.05 ; // Divertor resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"E_c2 = V_a - I_a2 * ( R_a + R_s ) - BD ; // BACK EMF in volt for I_a2\n",
+"E_c1 = V_a - I_a1 * ( R_a + R_s ) - BD ; // BACK EMF in volt for I_a1\n",
+"\n",
+"S_2 = S_1 * ( E_c2 / E_c1 ) * ( I_a1 / I_a2 ); // Speed in rpm at current I_a2\n",
+"\n",
+"E_c3 = V_a - I_a3 * ( R_a + R_s ) - BD ; // BACK EMF in volt for I_a3\n",
+"\n",
+"S_3 = S_1 * ( E_c3 / E_c1 ) * ( I_a1 / I_a3 ); // Speed in rpm at current I_a3\n",
+"\n",
+"// When divertor is connected in parallel to R_s\n",
+"R_sd = ( R_s * R_d ) / ( R_s + R_d ); // Effective series field resistance in ohm\n",
+"\n",
+"E_c2_new = V_a - I_a2 * ( R_a + R_sd ) - BD ; // BACK EMF in volt for I_a2\n",
+"S_2_new = S_1 * ( E_c2_new / E_c1 ) * ( I_a1 / ( I_a2 / 2 ) ); // Speed in rpm \n",
+"// at current I_a2\n",
+"\n",
+"E_c3_new = V_a - I_a3 * ( R_a + R_sd ) - BD ; // BACK EMF in volt for I_a3\n",
+"S_3_new = S_1 * ( E_c3_new / E_c1 ) * ( I_a1 / ( I_a3 / 2 ) ); // Speed in rpm \n",
+"// at current I_a3\n",
+"\n",
+"// Display the results\n",
+"disp(' Example 4-18 Solution : ');\n",
+"printf(' \n a: S_2 = %d rpm \n', S_2 );\n",
+"printf(' \n b: S_3 = %d rpm \n', S_3 );\n",
+"printf(' \n c: The effect of the divertor is to reduce the series field current');\n",
+"printf(' \n    (and flux) to half their previous values. ');\n",
+"printf(' \n    S_2 = %d rpm ', S_2_new );\n",
+"printf(' \n    S_3 = %d rpm \n', S_3_new );\n",
+"\n",
+"printf(' \n    The results may be tabulated as follows : \n ');\n",
+"printf(' \n     Case \t I_a in A \t S_o in rpm \t S_d in rpm ');\n",
+"printf(' \n    ________________________________________________________');\n",
+"printf(' \n      1    \t   %d       \t   %d         \t   ___ ', I_a1 , S_1  );\n",
+"printf(' \n      2.   \t   %d       \t   %d         \t   %d ', I_a2 , S_2 , S_2_new );\n",
+"printf(' \n      3.   \t   %d       \t   %d       \t   %d ', I_a3 , S_3 , S_3_new );\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.19: calculate_percentage_speed_regulation.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-19\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// From the calculations of Ex.4-16 , Ex.4-17 , Ex.4-18 we get no-load and\n",
+"// full-load speeds as follows\n",
+"S_n1 = 1810 ; // No-load speed in rpm (Ex.4-16)\n",
+"S_f1 = 1603 ; // Full-load speed in rpm (Ex.4-16)\n",
+"\n",
+"S_n2 = 1806 ; // No-load speed in rpm (Ex.4-17)\n",
+"S_f2 = 1231 ; // Full-load speed in rpm (Ex.4-17)\n",
+"\n",
+"S_n3 = 1311 ; // No-load speed in rpm (Ex.4-18)\n",
+"S_f3 = 505 ; // Full-load speed in rpm (Ex.4-18)\n",
+"\n",
+"// Calculations\n",
+"SR_1 = ( S_n1 - S_f1 ) / S_f1 * 100 ; // Speed regulation for shunt motor\n",
+"\n",
+"SR_2 = ( S_n2 - S_f2 ) / S_f2 * 100 ; // Speed regulation for compound motor\n",
+"\n",
+"SR_3 = ( S_n3 - S_f3 ) / S_f3 * 100 ; // Speed regulation for series motor\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-19 Solution : ');\n",
+"printf(' \n a: SR(shunt) = %.1f percent \n ', SR_1 );\n",
+"printf(' \n b: SR(compound) = %.1f percent \n ', SR_2 );\n",
+"printf(' \n c: SR(series) = %.1f percent \n ', SR_3 );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.1: calculate_force_and_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"d = 0.5; // diameter of the coil in m\n",
+"l = 0.6; // axial length of the coil in m\n",
+"B = 0.4; // flux density in T\n",
+"I = 25; // current carried by the coil in A\n",
+"theta = 60; // angle between the useful force & the interpolar ref axis in deg\n",
+"\n",
+"// Calculations\n",
+"F = B * I * l; // force developed on each coil side in  N\n",
+"f = F * sind(theta); // force developed at the instant the coil lies at an angle\n",
+"// of 60 w.r.t the interpolar  ref axis\n",
+"r = d / 2; // radius of the coil in m\n",
+"T_c = f * r; // torque developed in N-m\n",
+"T_c1 = T_c * 0.2248 * 3.281 ; // torque developed in lb-ft by first method\n",
+"T_c2 = T_c * 0.737562 ; // torque developed in lb-ft by second method\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-1 Solution : ')\n",
+"printf('\n a : F = %d N ', F );\n",
+"printf('\n b : f = %.2f N ', f );\n",
+"printf('\n c : Tc = %.2f N-m ', T_c );\n",
+"printf('\n d : 1.3 N-m * 0.2248 lb/N * 3.281 ft/m = %.2f lb-ft ', T_c1 );\n",
+"printf('\n     1.3 N-m * 0.737562 lb.ft/N.m = %.2f lb-ft ', T_c2 );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.20: calculate_no_load_speed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-20\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"SR = 0.1 ; // Given percent speed regulation 10% of a shunt motor \n",
+"omega_f1 = 60 * %pi ; // Full-load speed in rad/s\n",
+"\n",
+"// Calculations\n",
+"omega_n1 = omega_f1 * ( 1 + SR ); // No-load speed in rad/s\n",
+"\n",
+"S = omega_n1 * ( 1 / ( 2 * %pi )) * ( 60 / 1 ); // No-load speed in rpm \n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-20 Solution : ');\n",
+"printf(' \n a: omega_n1 = %.2f \n ', omega_n1);\n",
+"printf(' \n b: S = %d rpm ', S );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.21: calculate_internal_and_external_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-21\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"S_int = 1603 ; // Internal rated speed in rpm (Ex.4-16)\n",
+"S_ext = 1250 ; // External rated speed in rpm (Ex.4-16)\n",
+"hp_int = 14.3 ; // Internal horsepower\n",
+"hp_ext = 10 ; // External horsepower\n",
+"\n",
+"// Calculations\n",
+"T_int = ( hp_int * 5252 ) / S_int ; // Internal torque in lb-ft\n",
+"\n",
+"T_ext = ( hp_ext * 5252 ) / S_ext ; // External torque in lb-ft\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-21 Solution : ');\n",
+"printf(' \n a: T_int = %.2f lb-ft \n ', T_int );\n",
+"printf(' \n b: T_ext = %.2f lb-ft \n ', T_ext );\n",
+"printf(' \n c: Internal horsepower is developed as a result of electromagnetic');\n",
+"printf(' \n    torque produced by energy conversion. Some of the mechanical energy');\n",
+"printf(' \n    is used internally to overcome mechanical losses of the motor,');\n",
+"printf(' \n    reducing the torque available at its shaft to perform work.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.22: calculate_output_torque_in_ounceinches.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-22\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 50 ; // Power rating of the servo-motor in W\n",
+"S = 3000 ; // Full-load speed of the servo-motor in rpm\n",
+"\n",
+"// Calculation\n",
+"T_lbft = ( 7.04 * P ) / S ; // Output torque in lb-ft\n",
+"T_ounceinch = T_lbft * 192 ; // Output torque in ounce-inches\n",
+"\n",
+"// Display the result\n",
+"disp(' Example 4-22 Solution : ');\n",
+"printf(' \n  T = %.1f oz.in ', T_ounceinch );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.23: calculate_speed_and_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-23\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 50 ; // Power rating of the servo-motor in W\n",
+"S_rpm = 3000 ; // Full-load speed of the servo-motor in rpm\n",
+"\n",
+"// Calculations\n",
+"S_rad_per_sec = S_rpm * 2 * %pi / 60 ; // Full-load speed of the servo-motor \n",
+"// in rad/s\n",
+"omega = 314.2 ; // Angular frequency in rad/s\n",
+"T_Nm = P / omega ; // Output torque in Nm\n",
+"T_ounceinch = T_Nm * ( 1 / 7.0612e-3 ) ; // Output torque in oz.in\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-23 Solution : ');\n",
+"printf(' \n a: Speed in rad/s = %.1f rad/s \n ', S_rad_per_sec );\n",
+"printf(' \n b: T = %.4f N-m \n ', T_Nm );\n",
+"printf(' \n c: T = %.1f oz.in \n ', T_ounceinch );\n",
+"printf(' \n d: Both answers are the same.'); \n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.2: calculate_force_and_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"d = 18 ; // diameter of hte coil in inches\n",
+"l = 24 ; // axial length of the coil in inches\n",
+"B = 24000 ; // Flux density in lines/sq.inches\n",
+"I = 26 ; // Current carried by the coil in A\n",
+"theta = 60 ; // angle between the useful force & the interpolar ref axis in deg\n",
+"\n",
+"\n",
+"// Calculations\n",
+"F = ( B * I * l * 10 ^ -7 ) / 1.13 ; // force developed on each coil side in  lb\n",
+"f = F * sind(theta); // force developed at the instant the coil lies at an angle\n",
+"// of 60 w.r.t the interpolar  ref axis\n",
+"r = d / 2; // radius of the coil in inches\n",
+"T_c = f * ( r * 1 / 12); // torque developed in lb.ft/conductor\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-2 Solution : ')\n",
+"printf('\n a : F = %.3f lb ', F );\n",
+"printf('\n b : f = %.2f lb ', f );\n",
+"printf('\n c : Tc = %.3f lb-ft/conductor ', T_c );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.3: calculate_average_force_and_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"Z = 700 ; // no. of conductors\n",
+"d = 24 ; // diameter of the armature of the dc motor in inches\n",
+"l = 34 ; // axial length of the coil in inches\n",
+"B = 50000 ; // Flux density in lines/sq.inches\n",
+"I = 25 ; // Current carried by the coil in A\n",
+"\n",
+"// Calculations\n",
+"F_av = ( B * I * l * 10 ^ -7 ) / 1.13 * ( 700 * 0.7 ) ; // average force \n",
+"// developed on each coil side in  lb\n",
+"r = d / 2; // radius of the coil in inches\n",
+"T_av = F_av * ( r /12 ) ; // armature average torque in lb-ft\n",
+"\n",
+"// Display the results\n",
+"disp('Example 4-3 Solution : ')\n",
+"printf('\n a : Fav = %.2f lb ', F_av );\n",
+"printf('\n b : Tav = %.2f lb-ft ', T_av );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.4: calculate_torque_developed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"slots = 120 ; // No. of armature slots\n",
+"conductors_per_slot = 6 ; \n",
+"B = 60000 ; // Flux density in lines/sq.inches\n",
+"d = 28 ; // diameter of the armature\n",
+"l = 14 ; // axial length of the coil in inches\n",
+"A = 4 ; // No. of parallel paths\n",
+"span = 0.72 ; // Pole arcs span 72% of the armature surface\n",
+"I = 133.5 ; // Armature current in A\n",
+"\n",
+"// Calculations\n",
+"Z_Ta = slots * conductors_per_slot * span ; // No. of armature conductors\n",
+"F_t = ( B * I * l )/ ( 1.13 * 10 ^ 7  * A ) * Z_Ta ; // Force developed in lb\n",
+"r = ( d / 2 ) / 12 ; // radius of the armature in feet\n",
+"T = F_t * r ; // Tital torque developed\n",
+"\n",
+"// Display the result\n",
+"disp('Example 4-4 Solution : ')\n",
+"printf(' \n T = %d lb-ft', T );\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.5: calculate_armature_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"slots = 120 ; // No. of armature slots\n",
+"conductors_per_slot = 6 ; \n",
+"B = 60000 ; // Flux density in lines/sq.inches\n",
+"d = 28 ; // diameter of the armature\n",
+"l = 14 ; // axial length of the coil in inches\n",
+"A = 4 ; // No. of parallel paths\n",
+"span = 0.72 ; // Pole arcs span 72% of the armature surface\n",
+"T_a = 1500 ; // total armature torque in lb-ft\n",
+"\n",
+"// Calculation\n",
+"Z = slots * conductors_per_slot ; // No. of armature conductors\n",
+"r = ( d / 2 ) / 12 ; // radius of the armature in feet\n",
+"I_a = ( T_a * A * 1.13e7 ) / ( B * l * Z * r * span ) ; //Armature current in A\n",
+"\n",
+"// Display the result\n",
+"disp('Example 4-5 Solution : ')\n",
+"printf(' \n Ia = %.1f A ', I_a );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.6: calculate_torque_due_to_change_in_field_flux.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"T_old = 150 ; // Torque developed by a motor in N-m.\n",
+"disp('Example 4-6')\n",
+"disp('Given data : ')\n",
+"printf('\n \t\t\t phi \t I_a \t T ');\n",
+"printf('\n \t\t\t ________________________');\n",
+"printf('\n Original condition \t 1 \t 1 \t 150 N-m ');\n",
+"printf('\n New condition \t\t 0.9 \t 1.5 \t ? ');\n",
+"\n",
+"// Calculation\n",
+"T_new = T_old * ( 0.9 / 1 ) * ( 1.5 / 1 ) ; // New torque produced in N-m\n",
+"\n",
+"// Display the result\n",
+"printf('\n\n Solution : ')\n",
+"printf('\n Using the ratio method, the new torque is the product ');\n",
+"printf('\n of two new ratio changes : ');\n",
+"printf('\n T = %.1f N-m ', T_new );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.7: calculate_Ia_and_percentage_change_in_Ia_and_E.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"R_a = 0.25 ; // Armature resistance in ohm\n",
+"BD = 3 ; // Brush contact drop in volt\n",
+"V = 120 ; // Applied voltage in volt\n",
+"E_a = 110 ; // EMF in volt at a given load \n",
+"E_b = 105 ; // EMF in volt due to application of extra load \n",
+"\n",
+"// Calculations\n",
+"I_a_a = ( V - ( E_a + BD ) ) / R_a ; // Armature current for E_a\n",
+"I_a_b = ( V - ( E_b + BD ) ) / R_a ; // Armature current for E_b\n",
+"del_E = ( ( E_a - E_b ) / E_a ) * 100 ; // % change in counter EMF\n",
+"del_I = ( ( I_a_a - I_a_b ) / I_a_a ) * 100 ; // % change in armature current\n",
+"\n",
+"// Display the result\n",
+"disp('Example 4-7 Solution : ')\n",
+"printf('\n a : Ia = %d A ' , I_a_a );\n",
+"printf('\n b : At increased load \n     Ia = %d A ' , I_a_b );\n",
+"printf('\n c : del_Ec = %.2f percent \n     del_Ia = %.2f percent ' , del_E , del_I);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.8: calculate_speed_at_different_loads.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"V_a = 120 ; // Rated terminal voltage of the DC motor in volt\n",
+"R_a = 0.2 ; // Armature circuit resistance in ohm\n",
+"R_sh = 60 ; // Shunt field resistance in ohm\n",
+"I_l = 40 ; // Line current in A @ full load\n",
+"BD = 3 ; //  Brush voltage drop in volt\n",
+"S_orig = 1800 ; // Rated full-load speed in rpm\n",
+"\n",
+"// Calculations\n",
+"I_f = V_a / R_sh ; // Field current in A\n",
+"I_a = I_l - I_f ; // Armature current @ full load\n",
+"E_c_orig = V_a - ( I_a * R_a + BD ) ; // Back EMF @ full load\n",
+"\n",
+"I_a_a = I_a / 2 ; // Armature current @ half load\n",
+"E_c_a = V_a - ( I_a_a * R_a + BD ) ; // Back EMF @ half load\n",
+"S_a = S_orig * ( E_c_a / E_c_orig ) ; // Speed @ full load\n",
+"\n",
+"I_a_b = I_a * ( 5 / 4 ) ; // Armature current @ 125% overload\n",
+"E_c_b = V_a - ( I_a_b * R_a + BD ) ; // Back EMF @ 125% overload\n",
+"S_b = S_orig * ( E_c_b / E_c_orig ) ; // Speed @ 125% overload\n",
+"\n",
+"// Display the result\n",
+"disp('Example 4-8 Solution : ');\n",
+"\n",
+"printf(' \n a : At full load ');\n",
+"printf(' \n     S = %.1f rpm ' , S_a );\n",
+"\n",
+"printf(' \n b : At 125 percenytoverload ');\n",
+"printf(' \n     S = %.1f rpm ' , S_b );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 4.9: calculate_speed_with_increased_line_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 4: DC Dynamo Torque Relations-DC Motors\n",
+"// Example 4-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"I_l_orig = 40; // original Line current in A\n",
+"I_l_final = 66; // Final Line current in A\n",
+"\n",
+"phi_orig = 1; \n",
+"// field flux is increased by 12% so EMF produced and terminal \n",
+"// voltage will also increase by 12%\n",
+"phi_final = 1.12;\n",
+"\n",
+"V_a = 120;\n",
+"R_sh_orig = 60; // Original Field ckt resistance in ohm\n",
+"R_sh_final = 50 ; // Decreased final field ckt resistance in ohm\n",
+"\n",
+"R_a = 0.2; // Armature resistance in ohm\n",
+"BD = 3; // Brush voltage drop in volt\n",
+"S_orig = 1800; // Rated full-load speed\n",
+"\n",
+"// Calculations\n",
+"I_f_orig = V_a / R_sh_orig ; // Original Field current in A\n",
+"I_a_orig = I_l_orig - I_f_orig ; // Original Armature current @ full load\n",
+"E_c_orig = V_a - ( I_a_orig * R_a + BD ) ; // Back EMF @ full load\n",
+"\n",
+"I_f_final = V_a / R_sh_final ; // Final field current in A\n",
+"I_a_final = I_l_final - I_f_final ; // Final Armature current in A\n",
+"E_c_final = V_a - ( I_a_final * R_a + BD ) ; // Final EMF induced\n",
+"S = S_orig * ( E_c_final / E_c_orig ) * ( phi_orig / phi_final ) ; \n",
+"// Final speed of the motor\n",
+"\n",
+"// Display the result\n",
+"disp('Example 4-9 Solution : ');\n",
+"printf(' \n S = %.1f rpm ', S ); "
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/5-ARMATURE_REACTION_AND_COMMUTATION_IN_DYNAMOS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/5-ARMATURE_REACTION_AND_COMMUTATION_IN_DYNAMOS.ipynb
new file mode 100644
index 0000000..4e6fe1d
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/5-ARMATURE_REACTION_AND_COMMUTATION_IN_DYNAMOS.ipynb
@@ -0,0 +1,163 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 5: ARMATURE REACTION AND COMMUTATION IN DYNAMOS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.1: calculate_Zp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 5: ARMATURE REACTION AND COMMUTATION IN DYNAMOS\n",
+"// Example 5-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"conductors = 800 ; // No. of conductors \n",
+"I_a = 1000 ; // Rated armature current in A\n",
+"P = 10 ; // No. of poles\n",
+"pitch = 0.7 ; // Pole-face covers 70% of the pitch \n",
+"a = P ; // No. of parallel paths ( Simplex lap-wound )\n",
+"\n",
+"// Calculations\n",
+"// Using Eq.(5-1)\n",
+"Z = conductors / P ; // No. of armature conductors/path under each pole\n",
+"Z_a = Z * pitch ; // Active armature conductors/pole\n",
+"\n",
+"// Solving for Z_p using Z_p = Z_a / a\n",
+"Z_p = Z_a / a ; // No. of pole face conductors/pole \n",
+"\n",
+"// Display the results\n",
+"disp('Example 5-1 Solution : ');\n",
+"printf(' \n No. of pole face conductors/pole to give full armature reaction ');\n",
+"printf(' \n compensation, if the pole covers 70 percent of the pitch is : \n ');\n",
+"printf(' \n Z_p = %.1f conductors/pole ', Z_p );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.2: EX5_2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 5: ARMATURE REACTION AND COMMUTATION IN DYNAMOS\n",
+"// Example 5-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"conductors = 800 ; // No. of conductors \n",
+"I_a = 1000 ; // Rated armature current in A\n",
+"I_l = I_a ; // load or total current entering the armature in A\n",
+"P = 10 ; // No. of poles\n",
+"pitch = 0.7 ; // Pole-face covers 70% of the pitch \n",
+"a = P ; // No. of parallel paths ( Simplex lap-wound )\n",
+"alpha = 5 ; // No. of electrical degress that the brushes are shifted \n",
+"\n",
+"// Calculations\n",
+"Z = conductors / P ; // No. of armature conductors/path under each pole\n",
+"A_Z_per_pole = ( Z * I_l ) / ( P * a ); // Cross magnetizing\n",
+"// ampere-conductors/pole\n",
+"\n",
+"At_per_pole = ( 1 / 2 ) * ( 8000 / 1 ); // Ampere-turns/pole\n",
+"\n",
+"frac_demag_At_per_pole  = (2*alpha) / 180 * (At_per_pole); \n",
+"// Fraction of demagnetizing ampere-turns/pole\n",
+"\n",
+"funcprot(0); // to avoid redefining function: beta warning message\n",
+"\n",
+"beta = 180 - 2*alpha ; // cross-magnetizing electrical degrees\n",
+"\n",
+"cross_mag_At_per_pole = (beta/180)*(At_per_pole);\n",
+"// cross-magnetizing ampere-turns/pole\n",
+"\n",
+"// Display the results\n",
+"disp('Example 5-2 Solution : ');\n",
+"printf(' \n a: With the brushes on the GNA,the entire armature reaction effect');\n",
+"printf(' \n    is completely cross-magnetizing. The cross-magnetizing  ');\n",
+"printf(' \n    ampere-conductors/pole are ');\n",
+"printf(' \n    = %d ampere-conductots/pole \n', A_Z_per_pole);\n",
+"\n",
+"printf(' \n    and since there are 2 conductors/turn, the cross-magnetizing ');\n",
+"printf(' \n    ampere-turns/pole are \n    = %d At/pole \n\n', At_per_pole );\n",
+"\n",
+"\n",
+"printf(' \n b: Let alpha = the no. of electrical degrees that the brushes are ');\n",
+"printf(' \n    shifted. Then the total no. of demagnetizing electrical degrees ');\n",
+"printf(' \n    are 2*alpha, while the (remaining) cross-magnetizing electrical');\n",
+"printf(' \n    degrees,beta, are 180 - 2*alpha. The ratio of demagnetizing to ');\n",
+"printf(' \n    cross-magnetizing ampere-turns is always 2*alpha/beta. The ');\n",
+"printf(' \n    fraction of demagnetizing ampere-turns/pole is ');\n",
+"printf(' \n    = %.1f At/pole \n\n',frac_demag_At_per_pole );\n",
+"printf(' \n    Note: Slight calculation mistake in the textbook for case b\n')\n",
+"\n",
+"\n",
+"printf(' \n c: Since beta = 180-2*alpha = 170, the cross-magnetizing ampere-turns/pole ');\n",
+"printf(' \n    are \n    = %.1f At/pole ', cross_mag_At_per_pole );\n",
+"\n",
+"\n",
+"\n",
+"\n",
+"\n",
+"\n",
+""
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/6-AC_DYNAMO_VOLTAGE_RELATIONS_ALTERNATORS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/6-AC_DYNAMO_VOLTAGE_RELATIONS_ALTERNATORS.ipynb
new file mode 100644
index 0000000..9855493
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/6-AC_DYNAMO_VOLTAGE_RELATIONS_ALTERNATORS.ipynb
@@ -0,0 +1,675 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 6: AC DYNAMO VOLTAGE RELATIONS ALTERNATORS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.1: calculate_Eg_at_unity_PF_and_point75_lagging_PF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 1000 ; // kVA rating of the 3-phase alternator \n",
+"V_L = 4600 ; // Rated line voltage in volt\n",
+"// 3-phase, Y-connected alternator\n",
+"R_a = 2 ; // Armature resistance in ohm per phase\n",
+"X_s = 20 ; // Synchronous armature reactance in ohm per phase\n",
+"cos_theta_a = 1 ; // Unity power factor (case a)\n",
+"cos_theta_b = 0.75 ; // 0.75 power factor lagging (case b)\n",
+"sin_theta_b = sqrt( 1 - (cos_theta_b)^2 ); \n",
+"\n",
+"// Calculations\n",
+"V_P = V_L / sqrt(3) ; // Phase voltage in volt\n",
+"I_P = ( kVA * 1000 ) / ( 3*V_P ) ; // Phase current in A\n",
+"I_a  = I_P ; // Armature current in A\n",
+"\n",
+"// a: At unity  PF\n",
+"E_g_a = ( V_P + I_a * R_a ) + %i*(I_a*X_s);\n",
+"// Full-load generated voltage per-phase (case a)\n",
+"E_g_a_m=abs(E_g_a);//E_g_a_m=magnitude of E_g_a in volt\n",
+"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\n",
+"\n",
+"// b: At 0.75 PF lagging\n",
+"E_g_b = ( V_P*cos_theta_b + I_a * R_a ) + %i*( V_P*sin_theta_b + I_a*X_s );\n",
+"// Full-load generated voltage per-phase (case b )\n",
+"E_g_b_m=abs(E_g_b);//E_g_b_m=magnitude of E_g_b in volt\n",
+"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\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-1 Solution : ');\n",
+"printf('\n root 3 value is taken as %f , so slight variations in the answer.', sqrt(3));\n",
+"printf('\n\n a: At unity PF, \n ');\n",
+"printf('\n     Rectangular form :\n   E_g = '); disp(E_g_a);\n",
+"printf('\n     Polar form :');\n",
+"printf(' \n     E_g = %d <%.2f V/phase ', E_g_a_m , E_g_a_a );\n",
+"printf(' \n     where %d is magnitude and %.2f is phase angle\n',E_g_a_m,E_g_a_a);\n",
+"\n",
+"printf(' \n b: At 0.75 PF lagging , \n ');\n",
+"printf('\n     Rectangular form :\n   E_g = '); disp(E_g_b);\n",
+"printf('\n     Polar form :');\n",
+"printf(' \n     E_g = %d <%.2f V/phase ', E_g_b_m , E_g_b_a );\n",
+"printf(' \n     where %d is magnitude and %.2f is phase angle\n',E_g_b_m,E_g_b_a);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.2: calculate_Eg_at_point75_PF_and_point4_lead.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 1000 ; // kVA rating of the 3-phase alternator \n",
+"V_L = 4600 ; // Rated line voltage in volt\n",
+"// 3-phase, Y-connected alternator\n",
+"R_a = 2 ; // Armature resistance in ohm per phase\n",
+"X_s = 20 ; // Synchronous armature reactance in ohm per phase\n",
+"cos_theta_a = 0.75 ; // 0.75 PF leading (case a)\n",
+"cos_theta_b = 0.40 ; // 0.40 PF leading (case b)\n",
+"sin_theta_a = sqrt( 1 - (cos_theta_a)^2 ); // (case a)\n",
+"sin_theta_b = sqrt( 1 - (cos_theta_b)^2 ); // (case b)\n",
+"\n",
+"// Calculations\n",
+"V_P = V_L / sqrt(3) ; // Phase voltage in volt\n",
+"I_P = ( kVA * 1000 ) / ( 3*V_P ) ; // Phase current in A\n",
+"I_a  = I_P ; // Armature current in A\n",
+"\n",
+"// a: At 0.75 PF leading\n",
+"E_g_a = ( V_P*cos_theta_a + I_a * R_a ) + %i*( V_P*sin_theta_a - I_a*X_s);\n",
+"// Full-load generated voltage per-phase (case a)\n",
+"E_g_a_m=abs(E_g_a);//E_g_a_m=magnitude of E_g_a in volt\n",
+"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\n",
+"\n",
+"// b: At 0.40 PF leading\n",
+"E_g_b = ( V_P*cos_theta_b + I_a * R_a ) + %i*( V_P*sin_theta_b - I_a*X_s );\n",
+"// Full-load generated voltage per-phase (case b )\n",
+"E_g_b_m=abs(E_g_b);//E_g_b_m=magnitude of E_g_b in volt\n",
+"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\n",
+"\n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-2 Solution : ');\n",
+"printf('\n root 3 value is taken as %f , so slight variations in the answer.', sqrt(3));\n",
+"printf('\n\n a: 0.75 PF leading, \n ');\n",
+"printf('\n     Rectangular form :\n   E_g = '); disp(E_g_a);\n",
+"printf('\n     Polar form :');\n",
+"printf(' \n     E_g = %d <%.2f V/phase ', E_g_a_m , E_g_a_a );\n",
+"printf(' \n     where %d is magnitude and %.2f is phase angle\n',E_g_a_m,E_g_a_a);\n",
+"\n",
+"printf(' \n b: At 0.40 PF leading , \n ');\n",
+"printf('\n     Rectangular form :\n   E_g = '); disp(E_g_b);\n",
+"printf('\n     Polar form :');\n",
+"printf(' \n     E_g = %d <%.2f V/phase ', E_g_b_m , E_g_b_a );\n",
+"printf(' \n     where %d is magnitude and %.2f is phase angle\n',E_g_b_m,E_g_b_a);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.3: calculate_percent_voltage_regulation.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// From Ex.6-1 and Ex.6-2 we have V_P and E_g values as follows\n",
+"// Note : approximated values are considered when root 3 value is taken as 1.73 \n",
+"// as in textbook\n",
+"V_P = 2660 ; // Phase voltage \n",
+"E_g_a1 = 3836 ; // E_g at unity PF (Ex.6-1 case a)\n",
+"E_g_b1 = 4814 ; // E_g at 0.75 PF lagging (Ex.6-1 case b)\n",
+" \n",
+"E_g_a2 = 2364 ; // E_g at 0.75 PF leading (Ex.6-2 case a)\n",
+"E_g_b2 = 1315 ; // E_g at 0.40 PF leading (Ex.6-2 case b)\n",
+"\n",
+"// Calculations\n",
+"VR_a = ( E_g_a1 - V_P )/V_P * 100 ; // voltage regulation at unity PF  (Ex.6-1 case a)\n",
+"VR_b = ( E_g_b1 - V_P )/V_P * 100 ; // voltage regulation at 0.75 PF lagging (Ex.6-1 case b)\n",
+"\n",
+"VR_c = ( E_g_a2 - V_P )/V_P * 100 ; // voltage regulation at 0.75 PF leading (Ex.6-2 case a)\n",
+"VR_d = ( E_g_b2 - V_P )/V_P * 100 ; // voltage regulation at 0.40 PF leading (Ex.6-2 case b)\n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-3 Solution : ');\n",
+"printf(' \n a: At unity PF : ');\n",
+"printf(' \n    VR = %.1f percent \n ', VR_a );\n",
+"\n",
+"printf(' \n b: At 0.75 PF lagging : ');\n",
+"printf(' \n    VR = %.2f percent \n ', VR_b );\n",
+"\n",
+"printf(' \n c: At 0.75 PF leading : ');\n",
+"printf(' \n    VR = %.2f percent \n ', VR_c );\n",
+"\n",
+"printf(' \n d: At 0.40 PF leading : ');\n",
+"printf(' \n    VR = %.1f percent \n ', VR_d );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.4: calculate_Rdc_Rac_Zp_Xs_VR_at_point8_PF_lag_and_lead.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 100 ; // kVA rating of the 3-phase alternator\n",
+"V_L = 1100 ; // Line voltage of the 3-phase alternator in volt\n",
+"\n",
+"// dc-resistance test data\n",
+"E_gp1 = 6 ; // generated phase voltage in volt \n",
+"V_l = E_gp1 ; // generated line voltage in volt\n",
+"I_a1 = 10 ; // full-load current per phase in A\n",
+"cos_theta_b1 = 0.8 ; // 0.8 PF lagging (case b)\n",
+"cos_theta_b2 = 0.8 ; // 0.8 PF leading (case b)\n",
+"sin_theta_b1 = sqrt( 1 - (cos_theta_b1)^2 ); // (case b)\n",
+"sin_theta_b2 = sqrt( 1 - (cos_theta_b2)^2 ); // (case b)\n",
+"\n",
+"// open-circuit test data\n",
+"E_gp2 = 420 ; // generated phase voltage in volt \n",
+"I_f2 = 12.5 ; // Field current in A\n",
+"\n",
+"// short-circuit test data\n",
+"I_f3 = 12.5 ; // Field current in A\n",
+"// Line current I_l = rated value   in A\n",
+"\n",
+"// Calculations\n",
+"// Assuming that the alternator is Y-connected\n",
+"// case a :\n",
+"I_a_rated = (kVA*1000)/(V_L*sqrt(3)); // Rated current per phase in A\n",
+"I_a = sqrt(3)*I_a_rated ; // Rated Line current in A\n",
+" \n",
+"R_dc = V_l/(2*I_a1); // effective dc armature resistance in ohm/winding\n",
+"R_ac = R_dc * 1.5 ; // effective ac armature resistance in ohm.phase\n",
+"R_a = R_ac ; // effective ac armature resistance in ohm.phase from dc resistance test\n",
+"\n",
+"Z_p = E_gp2 / I_a ; // Synchronous impedance per phase\n",
+"X_s = sqrt( Z_p^2 - R_a^2 ); // Synchronous reactance per phase\n",
+"\n",
+"// case b :\n",
+"V_p = V_L / sqrt(3); // Phase voltage in volt (Y-connection)\n",
+"\n",
+"// At 0.8 PF lagging\n",
+"E_gp1 = ( V_p*cos_theta_b1 + I_a_rated * R_a ) + %i*( V_p*sin_theta_b1 + I_a_rated * X_s);\n",
+"E_gp1_m=abs(E_gp1);//E_gp1_m=magnitude of E_gp1 in volt\n",
+"E_gp1_a=atan(imag(E_gp1) /real(E_gp1))*180/%pi;//E_gp1_a=phase angle of E_gp1 in degrees\n",
+"V_n1 = E_gp1_m ; // No-load voltage in volt\n",
+"V_f1 = V_p ; // Full-load voltage in volt\n",
+"VR1 = ( V_n1 - V_f1 )/ V_f1 * 100; // percent voltage regulation at 0.8 PF lagging \n",
+"\n",
+"\n",
+"// At 0.8 PF leading\n",
+"E_gp2 = ( V_p*cos_theta_b2 + I_a_rated * R_a ) + %i*( V_p*sin_theta_b2 - I_a_rated*X_s);\n",
+"E_gp2_m=abs(E_gp2);//E_gp2_m=magnitude of E_gp2 in volt\n",
+"E_gp2_a=atan(imag(E_gp2) /real(E_gp2))*180/%pi;//E_gp2_a=phase angle of E_gp2 in degrees\n",
+"V_n2 = E_gp2_m ; // No-load voltage in volt\n",
+"V_f2 = V_p ; // Full-load voltage in volt\n",
+"VR2 = ( V_n2 - V_f2 )/V_f2 * 100 ; // percent voltage regulation at 0.8 PF leading \n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-4 Solution : ');\n",
+"printf(' \n Assuming that the alternator is Y-connected ');\n",
+"printf(' \n a: R_dc = %.1f ohm/winding ', R_dc );\n",
+"printf(' \n    R_ac = %.2f ohm/phase ', R_ac );\n",
+"printf(' \n    Z_p = %.2f ohm/phase ', Z_p );\n",
+"printf(' \n    X_s = %.2f ohm/phase \n', X_s );\n",
+"\n",
+"printf(' \n b: At 0.8 PF lagging ');\n",
+"printf(' \n    Percent voltage regulation = %.1f percent \n', VR1 );\n",
+"\n",
+"printf(' \n    At 0.8 PF leading ');\n",
+"printf(' \n    Percent voltage regulation = %.1f percent ', VR2 );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.5: calculate_prev_eg_values_for_delta_connection.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 100 ; // kVA rating of the 3-phase alternator\n",
+"V_L = 1100 ; // Line voltage of the 3-phase alternator in volt\n",
+"\n",
+"// dc-resistance test data\n",
+"E_gp1 = 6 ; // generated phase voltage in volt \n",
+"V_l = E_gp1 ; // generated line voltage in volt\n",
+"I_a1 = 10 ; // full-load current per phase in A\n",
+"cos_theta_b1 = 0.8 ; // 0.8 PF lagging (case b)\n",
+"cos_theta_b2 = 0.8 ; // 0.8 PF leading (case b)\n",
+"sin_theta_b1 = sqrt( 1 - (cos_theta_b1)^2 ); // (case b)\n",
+"sin_theta_b2 = sqrt( 1 - (cos_theta_b2)^2 ); // (case b)\n",
+"\n",
+"// open-circuit test data\n",
+"E_gp2 = 420 ; // generated phase voltage in volt \n",
+"I_f2 = 12.5 ; // Field current in A\n",
+"\n",
+"// short-circuit test data\n",
+"I_f3 = 12.5 ; // Field current in A\n",
+"// Line current I_l = rated value   in A\n",
+"\n",
+"// Calculations\n",
+"// Assuming that the alternator is delta-connected\n",
+"// case a :\n",
+"I_a_rated = (kVA*1000)/(V_L*sqrt(3)); // Rated current per phase in A\n",
+"I_L = I_a_rated ; // Line current in A \n",
+"\n",
+"V_p = E_gp2 ; // Phase voltage in volt\n",
+"V_l = V_p ; // Line voltage in volt (from short circuit data)\n",
+"\n",
+"I_p = I_L / sqrt(3) ; // Phase current in A (delta connection)\n",
+"I_a = I_p ; // Rated current in A\n",
+"\n",
+"Z_s = V_l / I_p ; // Synchronous impedance per phase\n",
+"R_dc = E_gp1/(2*I_a1); // effective dc armature resistance in ohm/winding\n",
+"R_ac = R_dc * 1.5 ; // effective ac armature resistance in ohm.phase\n",
+"\n",
+"// R_eff in delta = 3 * R_eff in Y\n",
+"R_eff = 3 * R_ac ; // Effective armature resistance in ohm\n",
+"R_a = R_eff ; // effective ac armature resistance in ohm.phase from dc resistance test\n",
+"\n",
+"X_s = sqrt( Z_s^2 - R_a^2 ); // Synchronous reactance per phase\n",
+"\n",
+"V_p = V_L ; // Phase voltage in volt (delta-connection)\n",
+"\n",
+"// At 0.8 PF lagging\n",
+"E_gp1 = ( V_p*cos_theta_b1 + I_a * R_a ) + %i*( V_p*sin_theta_b1 + I_a*X_s);\n",
+"E_gp1_m=abs(E_gp1);//E_gp1_m=magnitude of E_gp1 in volt\n",
+"E_gp1_a=atan(imag(E_gp1) /real(E_gp1))*180/%pi;//E_gp1_a=phase angle of E_gp1 in degrees\n",
+"V_n1 = E_gp1_m ; // No-load voltage in volt\n",
+"V_f1 = V_p ; // Full-load voltage in volt\n",
+"VR1 = ( V_n1 - V_f1 )/ V_f1 * 100; // percent voltage regulation at 0.8 PF lagging \n",
+"\n",
+"\n",
+"// At 0.8 PF leading\n",
+"E_gp2 = ( V_p*cos_theta_b2 + I_a * R_a ) + %i*( V_p*sin_theta_b2 - I_a*X_s);\n",
+"E_gp2_m=abs(E_gp2);//E_gp2_m=magnitude of E_gp2 in volt\n",
+"E_gp2_a=atan(imag(E_gp2) /real(E_gp2))*180/%pi;//E_gp2_a=phase angle of E_gp2 in degrees\n",
+"V_n2 = E_gp2_m ; // No-load voltage in volt\n",
+"V_f2 = V_p ; // Full-load voltage in volt\n",
+"VR2 = ( V_n2 - V_f2 )/V_f2 * 100 ; // percent voltage regulation at 0.8 PF leading \n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-5 Solution : ');\n",
+"printf(' \n Assuming that the alternator is delta-connected : \n ');\n",
+"printf(' \n a: I_p = %.3f A ', I_p );\n",
+"printf(' \n    Z_s = %.2f ohm/phase ', Z_s );\n",
+"printf(' \n    R_eff in delta = %.2f ohm/phase ', R_eff );\n",
+"printf(' \n    X_s = %.1f ohm/phase \n', X_s );\n",
+"printf(' \n    R_eff, reactance and impedance per phase in delta is 3 times')\n",
+"printf(' \n    the value when connected in Y. \n')\n",
+"\n",
+"printf(' \n b: At 0.8 PF lagging ');\n",
+"printf(' \n    Percent voltage regulation = %.1f percent \n', VR1 );\n",
+"\n",
+"printf(' \n    At 0.8 PF leading ');\n",
+"printf(' \n    Percent voltage regulation = %.1f percent \n', VR2 );\n",
+"printf(' \n    Percentage voltage regulation remains the same both in Y and delta connection.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.6: calculate_Imax_overload_and_Isteady.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3-phase Y-connected alternator\n",
+"E_L = 11000 ; // Line voltage generated in volt\n",
+"kVA = 165000 ; // kVA rating of the alternator\n",
+"R_p = 0.1 ; // Armature resistance in ohm/per phase\n",
+"Z_p = 1.0 ; // Synchronous reactance/phase\n",
+"Z_r = 0.8 ; // Reactor reactance/phase\n",
+"\n",
+"// Calculations\n",
+"E_p = E_L / sqrt(3); // Rated phase voltage in volt\n",
+"I_p = (kVA * 1000)/(3*E_p); // Rated current per phase in A\n",
+"\n",
+"// case a\n",
+"I_max_a = E_p / R_p ; // Maximum short-circuit current in A (case a)\n",
+"overload_a = I_max_a / I_p ; // Overload (case a)\n",
+"\n",
+"// case b\n",
+"I_steady = E_p / Z_p ; // Sustained short-circuit current in A\n",
+"overload_b = I_steady / I_p ; // Overload (case b)\n",
+"\n",
+"// case c\n",
+"Z_t = R_p + %i*Z_r ; // Total reactance per phase\n",
+"I_max_c = E_p / Z_t ; // Maximum short-circuit current in A (case b)\n",
+"I_max_c_m=abs(I_max_c);//I_max_c_m=magnitude of I_max_c in A\n",
+"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\n",
+"overload_c = I_max_c_m / I_p ; // Overload (case a)\n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-6 Solution : ');\n",
+"printf('\n root 3 value is taken as %f , so slight variations in the answer.\n', sqrt(3));\n",
+"printf(' \n a: I_max = %d A ', I_max_a );\n",
+"printf(' \n    overload = %.1f * rated current \n', overload_a );\n",
+"\n",
+"printf(' \n b: I_steady = %d A ', I_steady );\n",
+"printf(' \n    overload = %.2f * rated current \n', overload_b );\n",
+"\n",
+"printf(' \n c: Rectangular form :\n   I_max = '); disp(I_max_c);\n",
+"printf(' \n    Polar form :');\n",
+"printf(' \n    I_max = %d <%.2f A ', I_max_c_m , I_max_c_a );\n",
+"printf(' \n    where %d is magnitude and %.2f is phase angle\n',I_max_c_m,I_max_c_a);\n",
+"printf(' \n    overload = %.3f * rated current \n', overload_c );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.7: calculate_P_and_Pperphase_and_Egp_magnitude_phase_angle_and_torque_angle.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA = 100 ; // kVA rating of the 3-phase alternator\n",
+"V_L = 1100 ; // Line voltage of the 3-phase alternator in volt\n",
+"\n",
+"// dc-resistance test data\n",
+"E_gp1 = 6 ; // generated phase voltage in volt \n",
+"V_l = E_gp1 ; // generated line voltage in volt\n",
+"I_a1 = 10 ; // full-load current per phase in A\n",
+"cos_theta = 0.8 ; // 0.8 PF lagging \n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 ); // \n",
+"\n",
+"// open-circuit test data\n",
+"E_gp2 = 420 ; // generated phase voltage in volt \n",
+"I_f2 = 12.5 ; // Field current in A\n",
+"\n",
+"// short-circuit test data\n",
+"I_f3 = 12.5 ; // Field current in A\n",
+"// Line current I_l = rated value   in A\n",
+"\n",
+"// Calculated data from Ex.6-4\n",
+"I_L = 52.5 ; // Rated line current in A\n",
+"I_a = I_L ; // Rated current per phase in A\n",
+"E_gp = 532 + %i*623 ; // Generated voltage at 0.8 PF lagging\n",
+"X_s = 4.6 ; // Synchronous reactance per phase\n",
+"V_p = 635 ; // Phase voltage in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"P_T = sqrt(3) * V_L * I_L * cos_theta ; // Total output 3-phase power\n",
+"\n",
+"// case b\n",
+"P_p_b = P_T / 3 ; // Total output 3-phase power per phase\n",
+"\n",
+"// case c\n",
+"E_gp_m=abs(E_gp);//E_gp_m=magnitude of E_gp in volt\n",
+"E_gp_a=atan(imag(E_gp) /real(E_gp))*180/%pi;//E_gp_a=phase angle of E_gp in degrees\n",
+"\n",
+"// case d\n",
+"theta = acos(0.8)*180/%pi; // phase angle for PF in degrees\n",
+"theta_plus_deba = E_gp_a ; // phase angle of E_gp in degrees\n",
+"deba = theta_plus_deba - theta ; // Torque angle in degrees\n",
+"\n",
+"// case e\n",
+"P_p_e = (E_gp_m/X_s)*V_p*sind(deba); // Approximate output power/phase (Eq.(6-10))\n",
+"\n",
+"// case f\n",
+"P_p_f = E_gp_m * I_a * cosd(theta_plus_deba); // Approximate output power/phase (Eq.(6-9))\n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-7 Solution : ');\n",
+"printf('\n root 3 value is taken as %f , so slight variations in the answer.\n', sqrt(3));\n",
+"printf(' \n a: P_T = %d W \n', P_T );\n",
+"printf(' \n b: P_p = %.2f W \n', P_p_b );\n",
+"printf(' \n c: E_gp = %d <%.2f V \n', E_gp_m, E_gp_a );\n",
+"printf(' \n    where %d is magnitude in V and %.2f is phase angle in degrees.\n',E_gp_m,E_gp_a);\n",
+"printf(' \n d: Torque angle, deba = %.2f degrees \n', deba );\n",
+"printf(' \n e: P_p = %d W \n', P_p_e );\n",
+"printf(' \n f: P_p = %d W ', P_p_f );\n",
+"\n",
+" "
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 6.8: calculate_torqueperphase_and_total_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS\n",
+"// Example 6-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"\n",
+"kVA = 100 ; // kVA rating of the 3-phase alternator\n",
+"V_L = 1100 ; // Line voltage of the 3-phase alternator in volt\n",
+"S = 1200 ; // Synchronous speed in rpm\n",
+"\n",
+"// dc-resistance test data\n",
+"E_gp1 = 6 ; // generated phase voltage in volt \n",
+"V_l = E_gp1 ; // generated line voltage in volt\n",
+"I_a1 = 10 ; // full-load current per phase in A\n",
+"cos_theta = 0.8 ; // 0.8 PF lagging \n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 ); // \n",
+"\n",
+"// open-circuit test data\n",
+"E_gp2 = 420 ; // generated phase voltage in volt \n",
+"I_f2 = 12.5 ; // Field current in A\n",
+"\n",
+"// short-circuit test data\n",
+"I_f3 = 12.5 ; // Field current in A\n",
+"// Line current I_l = rated value   in A\n",
+"\n",
+"// Calculated data from Ex.6-4 & Ex.6-7\n",
+"I_L = 52.5 ; // Rated line current in A\n",
+"I_a = I_L ; // Rated current per phase in A\n",
+"E_gp = 532 + %i*623 ; // Generated voltage at 0.8 PF lagging\n",
+"E_g = 819 ; // E_g = magnitude of E_gp in volt\n",
+"X_s = 4.6 ; // Synchronous reactance per phase\n",
+"V_p = 635 ; // Phase voltage in volt\n",
+"deba = 12.63 ; // Torque angle in degrees\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"T_p_a  = ( 7.04 * E_g * V_p * sind(deba) ) / (S*X_s); // Output torque per phase in lb.ft\n",
+"T_3phase_a = 3 * T_p_a ; // Output torque for 3-phase in lb.ft\n",
+"\n",
+"// case b\n",
+"omega = S * 2*%pi *(1/60); // Angular frequency in rad/s\n",
+"T_p_b = ( E_g * V_p * sind(deba))/(omega*X_s); // Output torque per phase in lb.ft\n",
+"T_3phase_b = 3 * T_p_b ; // Output torque for 3-phase in lb.ft\n",
+"\n",
+"// case c\n",
+"T_p_c = T_p_a * 1.356 ; // Output torque per phase in N.m\n",
+"T_3phase_c = 3 * T_p_c ; // Output torque for 3-phase in N.m\n",
+"\n",
+"// Display the results\n",
+"disp('Example 6-8 Solution : ');\n",
+"pi = %pi;\n",
+"printf(' \n    Slight variations in the answers are due to value of pi = %f ',pi);\n",
+"printf(' \n    and omega = %f, which are slightly different as in the textbook.\n',omega);\n",
+"printf(' \n a: T_p = %d lb-ft ',T_p_a);\n",
+"printf(' \n    T_3phase = %d lb-ft \n', T_3phase_a);\n",
+"\n",
+"printf(' \n b: T_p = %.1f N-m ',T_p_b);\n",
+"printf(' \n    T_3phase = %.1f N-m \n', T_3phase_b);\n",
+"\n",
+"printf(' \n c: T_p = %.1f N-m ',T_p_c);\n",
+"printf(' \n    T_3phase = %.1f N-m \n', T_3phase_c);\n",
+"printf(' \n    Answers from cases b and c almost tally each other ');"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/7-PARALLEL_OPERATION.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/7-PARALLEL_OPERATION.ipynb
new file mode 100644
index 0000000..83952f9
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/7-PARALLEL_OPERATION.ipynb
@@ -0,0 +1,880 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 7: PARALLEL OPERATION"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.10: calculate_synchronizing_I_and_P_and_P_losses.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"E_2_mag = 230 ; // Magnitude of voltage generated by alternator 2 in volt\n",
+"E_1_mag = 230 ; // Magnitude of voltage generated by alternator 1 in volt\n",
+"\n",
+"theta_2 = 180 ; // Phase angle of generated voltage by alternator 2 in degrees\n",
+"theta_1 = 20 ; // Phase angle of generated voltage by alternator 1 in degrees\n",
+"\n",
+"// writing given voltage in exponential form as follows\n",
+"// %pi/180 for degrees to radians conversion\n",
+"E_2 = E_2_mag * expm(%i * theta_2*(%pi/180) ); // voltage generated by alternator 2 in volt\n",
+"E_1 = E_1_mag * expm(%i * theta_1*(%pi/180) ); // voltage generated by alternator 1 in volt\n",
+"\n",
+"// writing given impedance(in ohm)in exponential form as follows\n",
+"Z_1 = 6 * expm(%i * 50*(%pi/180) ); // %pi/180 for degrees to radians conversion\n",
+"Z_2 = Z_1 ;\n",
+"Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees\n",
+"\n",
+"// Calculations\n",
+"E_r = E_2 + E_1 ; // Total voltage generated by Alternator 1 and 2 in volt\n",
+"E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt\n",
+"E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees\n",
+"\n",
+"// case a\n",
+"I_s = E_r / (Z_1 + Z_2); // Synchronozing current in A\n",
+"I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A\n",
+"I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees\n",
+"\n",
+"// case b\n",
+"E_gp1 = E_1_mag;\n",
+"P_1 = E_gp1 * I_s_m * cosd(I_s_a - theta_1); // Synchronozing power developed by alternator 1 in W\n",
+"\n",
+"// case c\n",
+"E_gp2 = E_2_mag;\n",
+"P_2 = E_gp2 * I_s_m * cosd(I_s_a - theta_2); // Synchronozing power developed by alternator 2 in W\n",
+"\n",
+"// case d\n",
+"// but consider +ve vlaue for P_2 for finding losses, so\n",
+"P2 = abs(P_2);\n",
+"losses = P_1 - P2 ; // Losses in the armature in W\n",
+"\n",
+"// E_r_a yields -80 degrees which is equivalent to 100 degrees, so\n",
+"theta = 100 - I_s_a ; // Phase difference between E_r and I_s in degrees\n",
+"\n",
+"check = E_r_m * I_s_m * cosd(theta); // Verifying losses by Eq.7-7\n",
+"R_aT = 12*cosd(50) ; // total armature resistance of alternator 1 and 2 in ohm\n",
+"double_check = (I_s_m)^2 * (R_aT); // Verifying losses by Eq.7-7\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-10 Solution : ');\n",
+"printf(' \n a: I_s = ');disp(I_s);\n",
+"printf(' \n    I_s = %.2f <%.2f A \n ',I_s_m, I_s_a );\n",
+"\n",
+"printf(' \n b: P_1 = %.f W (power delivered to bus)',P_1);\n",
+"printf(' \n    Note:Slight variation in P_1 is due slight variations in ')\n",
+"printf(' \n         phase angle of I_s,& angle btw (E_gp1,I_s)\n')\n",
+"printf(' \n    P_2 = %.f W (power received from bus)\n',P_2);\n",
+"\n",
+"printf(' \n c: Losses: P_1 - P_2 = %.f W',losses);\n",
+"printf(' \n    Check: E_a*I_s*cos(theta) = %.f W ',check );\n",
+"printf(' \n    Double check : (I_s)^2*(R_a1+R_a2) = %.f W ',double_check );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.11: calculate_mesh_currents_line_currents_phase_voltages_phasor_diagram.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// writing supply voltage in exponential form as follows\n",
+"// %pi/180 for degrees to radians conversion\n",
+"V_AB = 100 * expm(%i * 0*(%pi/180) ); // voltage supplied across A & B in volt\n",
+"V_BC = 100 * expm(%i * -120*(%pi/180) ); // voltage supplied across B & C in volt\n",
+"V_CA = 100 * expm(%i * 120*(%pi/180) ); // voltage supplied across C & A in volt\n",
+"\n",
+"disp('Example 7-11 : ');\n",
+"printf('\n Writing two mesh equations for I_1 and I_2 in fig.7-23a yields following\n array :'); \n",
+"printf(' \n  I_1 \t\t I_2 \t\t V ');\n",
+"printf(' \n ____________________________________________');\n",
+"printf(' \n  6 + j0 \t -3 + j0 \t 100 + j0 ');\n",
+"printf(' \n -3 + j0 \t 3 - j4 \t -50 - j86.6 ');\n",
+"\n",
+"// Calculations\n",
+"A = [ (6+%i*0) (-3+%i*0) ; (-3+%i*0) (3-%i*4) ]; // Matrix containing above mesh eqns array\n",
+"delta = det(A); // Determinant of A\n",
+"\n",
+"// case a \n",
+"I_1 = det( [ (100+%i*0) (-3+%i*0) ; (-50-%i*86.60) (3-%i*4) ] ) / delta ;\n",
+"// Mesh current I_1 in A \n",
+"I_1_m = abs(I_1);//I_1_m=magnitude of I_1 in A\n",
+"I_1_a = atan(imag(I_1) /real(I_1))*180/%pi;//I_1_a=phase angle of I_1 in degrees\n",
+"\n",
+"I_2 = det( [ (6+%i*0) (100+%i*0) ; (-3+%i*0) (-50-%i*86.6) ] ) / delta ;\n",
+"// Mesh current I_2 in A \n",
+"I_2_m = abs(I_2);//I_2_m=magnitude of I_2 in A\n",
+"I_2_a = atan(imag(I_2) /real(I_2))*180/%pi;//I_2_a=phase angle of I_2 in degrees\n",
+"\n",
+"// case b\n",
+"I_A = I_1 ; //  Line current I_A in A\n",
+"I_A_m = abs(I_A);//I_A_m=magnitude of I_A in A\n",
+"I_A_a = atan(imag(I_A) /real(I_A))*180/%pi;//I_A_a=phase angle of I_A in degrees\n",
+"\n",
+"I_B = I_2 - I_1 ; //  Line current I_B in A\n",
+"I_B_m = abs(I_B);//I_B_m=magnitude of I_B in A\n",
+"I_B_a = atan(imag(I_B) /real(I_B))*180/%pi - 180;//I_B_a=phase angle of I_B in degrees\n",
+"\n",
+"I_C = -I_2 ; //  Line current I_C in A\n",
+"I_C_m = abs(I_C);//I_C_m=magnitude of I_C in A\n",
+"I_C_a = 180 + atan(imag(I_C) /real(I_C))*180/%pi;//I_C_a=phase angle of I_C in degrees\n",
+"\n",
+"// case c\n",
+"Z_A = 3 * expm(%i * 0*(%pi/180) ); // Impedance in line A in ohm\n",
+"Z_B = 3 * expm(%i * 0*(%pi/180) ); // Impedance in line B in ohm\n",
+"Z_C = 4 * expm(%i * -90*(%pi/180) ); // Impedance in line C in ohm\n",
+"\n",
+"V_AO = I_A * Z_A ; // Phase voltage V_AO in volt\n",
+"V_AO_m = abs(V_AO);//V_AO_m=magnitude of V_AO in volt\n",
+"V_AO_a = atan(imag(V_AO) /real(V_AO))*180/%pi;//V_AO_a=phase angle of V_AO in degrees\n",
+"\n",
+"V_BO = I_B * Z_B ; // Phase voltage V_BO in volt\n",
+"V_BO_m = abs(V_BO);//V_BO_m=magnitude of V_BO in volt\n",
+"V_BO_a = atan(imag(V_BO) /real(V_BO))*180/%pi - 180;//V_BO_a=phase angle of V_BO in degrees\n",
+"\n",
+"V_CO = I_C * Z_C ; // Phase voltage V_CO in volt\n",
+"V_CO_m = abs(V_CO);//V_CO_m=magnitude of V_CO in volt\n",
+"V_CO_a = atan(imag(V_CO) /real(V_CO))*180/%pi;//V_CO_a=phase angle of V_CO in degrees\n",
+"\n",
+"// Display the results\n",
+"disp('Solution : ');\n",
+"printf(' \n a: I_1 in A = ');disp(I_1);\n",
+"printf(' \n    I_1 = %.2f <%.2f A \n ',I_1_m, I_1_a );\n",
+"printf(' \n    I_2 in A = ');disp(I_2);\n",
+"printf(' \n    I_2 = %.2f <%.2f A\n ',I_2_m, I_2_a );\n",
+"\n",
+"printf(' \n b: I_A in A = ');disp(I_1);\n",
+"printf(' \n    I_A = %.2f <%.2f A\n',I_A_m, I_A_a );\n",
+"\n",
+"printf(' \n    I_B in A = ');disp(I_B);\n",
+"printf(' \n    I_B = %.2f <%.2f A\n',I_B_m, I_B_a );\n",
+"\n",
+"printf(' \n    I_C in A = ');disp(I_C);\n",
+"printf(' \n    I_C = %.2f <%.2f A \n',I_C_m,  I_C_a );\n",
+"\n",
+"printf(' \n c: V_AO = %.2f <%.2f V',V_AO_m, V_AO_a );\n",
+"printf(' \n    V_BO = %.2f <%.2f V',V_BO_m, V_BO_a );\n",
+"printf(' \n    V_CO = %.2f <%.2f V\n',V_CO_m, V_CO_a );\n",
+"\n",
+"printf(' \n d: The phasor diagram is shown in Fig.7-23b, with the phase voltages');\n",
+"printf(' \n    inscribed inside the (equilateral) triangle of given line voltages');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.1: calculate_I_Ia_and_P.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"R_sh = 120 ; // Shunt field resistance in ohm\n",
+"R_a = 0.1 ; // Armature resistance in ohm\n",
+"V_L = 120 ; // Line voltage in volt\n",
+"E_g1 = 125 ; // Generated voltage by dynamo A\n",
+"E_g2 = 120 ; // Generated voltage by dynamo B\n",
+"E_g3 = 114 ; // Generated voltage by dynamo C\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// 1:\n",
+"I_gA = ( E_g1 - V_L ) / R_a ; // Current in the generating source A ( in  A)\n",
+"I_f = V_L / R_sh ; // Shunt field current in A\n",
+"I_a1 = I_gA + I_f ; // Armature current in A for generator A\n",
+"I_L1 = I_gA ; // Current delivered by dynamo A to the bus in A\n",
+"\n",
+"// 2:\n",
+"I_gB =  ( E_g2 - V_L ) / R_a ; // Current in the generating source B ( in  A)\n",
+"I_a2 = I_gB + I_f ; // Armature current in A for generator B\n",
+"I_L2 = I_gB ; // Current delivered by dynamo B to the bus in A\n",
+"\n",
+"// 3:\n",
+"I_gC =  ( V_L - E_g3 ) / R_a ; // Current in the generating source C ( in  A)\n",
+"I_a3 = I_gC ; // Armature current in A for generator C\n",
+"I_L3 = I_gC + I_f ; // Current delivered by dynamo C to the bus in A\n",
+"\n",
+"// case b\n",
+"// 1:\n",
+"P_LA = V_L * I_L1 ; // Power delivered to the bus by dynamo A in W\n",
+"P_gA = E_g1 * I_a1 ; // Power generated by dynamo A\n",
+"\n",
+"// 2:\n",
+"P_LB = V_L * I_L2 ; // Power delivered to the bus by dynamo B in W\n",
+"P_gB = E_g2 * I_a2 ; // Power generated by dynamo B\n",
+"\n",
+"// 3:\n",
+"P_LC = V_L * I_L3 ; // Power delivered to the bus by dynamo C in W\n",
+"P_gC = E_g3 * I_a3 ; // Power generated by dynamo C\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-1 Solution : ');\n",
+"printf(' \n a: 1. I_gA = %d A \t I_f = %d A ', I_gA,I_f );\n",
+"printf(' \n       Thus,dynamo A delivers %d A to the bus and has an armature', I_gA);\n",
+"printf(' \n       current of %d A + %d A = %d \n', I_gA,I_f,I_a1 );\n",
+"\n",
+"printf(' \n    2. I_gB = %d A ', I_gB);\n",
+"printf(' \n       Thus, dynamo B is floating and has as armature & field current of %d A \n',I_f);\n",
+"\n",
+"printf(' \n    3. I_gC = %d A ',I_gC);\n",
+"printf(' \n       Dynamo C receives %d A from the bus & has an armature current of %d A\n',I_L3,I_a3);\n",
+"\n",
+"printf(' \n b: 1. Power delivered to the bus by dynamo A is : ');\n",
+"printf(' \n       P_LA = %d W ',P_LA);\n",
+"printf(' \n       Power generated by dynamo A is \n       P_gA = %d W \n',P_gA);\n",
+" \n",
+"printf(' \n    2. Since dynamo B neither delivers power to nor receives power from the bus, ');\n",
+"printf(' \n       P_B = %d W ',P_LB);\n",
+"printf(' \n       Power generated by dynamo B,to excite its field, is');\n",
+"printf(' \n       P_gB = %d W \n ', P_gB); \n",
+"\n",
+"printf(' \n    3. Power delivered by the bus to dynamo C is ');\n",
+"printf(' \n       P_LC = %d W ', P_LC);\n",
+"printf(' \n       while the internal power delivered in the direction of rotation');\n",
+"printf(' \n       of its prime mover to aid rotation is \n       P_gC = %d W', P_gC );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.2: calculate_all_currents_and_power_of_the_generator.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-2\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"R_a = 0.1 ; // Armature resistance in ohm\n",
+"R_f = 100 ; // Field ckt resistance in ohm\n",
+"V_L_b = 120 ; // Bus voltage in volt\n",
+"V_L_a = 140 ; // Voltage of the generator in volt\n",
+"V_f = V_L_a ; // Voltage across the field in volt\n",
+"// Calculations\n",
+"// case a\n",
+"I_f_a = V_f / R_f ; // Field current in A\n",
+"I_a_a = I_f_a ; // Armature current in A\n",
+"E_g_a = V_L_a + I_a_a * R_a ; // Generated EMF in volt\n",
+"P_g_a = E_g_a * I_a_a ; // Generated power in W\n",
+"// case b\n",
+"I_a_b = ( E_g_a - V_L_b ) / R_a ; // Armature current in A\n",
+"I_f_b = V_L_b / R_f ; // Field current in A\n",
+"I_Lg = I_a_b - I_f_b ; // Generated line current in A\n",
+"P_L = V_L_b * I_Lg ; // Power generated across the lines in W\n",
+"E_g_b = V_L_a ;\n",
+"P_g_b = E_g_b * I_a_b ; // Generated power in W\n",
+"// Display the results \n",
+"disp('Example 7-2 Solution : ');\n",
+"printf(' \n a: Before it is connected to the bus ');\n",
+"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);\n",
+"printf(' \n b: After it is connected to the bus ');\n",
+"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 );\n",
+"printf(' \n    P_L = %.f W \n    P_g = %.f W ', P_L , P_g_b );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.3: calculate_VL_IL_Pg_and_PL.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"R_a = 0.1 ; // Armature resistance in ohm of 3 shunt generators\n",
+"R_a1 =R_a ; \n",
+"R_a2 =R_a ; \n",
+"R_a3 =R_a ; \n",
+"R_L = 2 ; // Load resistance in ohm\n",
+"E_g1 = 127 ; // Voltage generated by generator 1 in volt\n",
+"E_g2 = 120 ; // Voltage generated by generator 2 in volt\n",
+"E_g3 = 119 ; // Voltage generated by generator 3 in volt\n",
+"// Neglect field currents\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// Terminal bus voltage in volt\n",
+"V_L = ( (127/0.1) + (120/0.1) + (119/0.1) ) / ( (1/0.1) + (1/0.1) + (1/0.1) + 0.5);\n",
+"\n",
+"// case b\n",
+"I_L1 = (E_g1 - V_L)/R_a1 ; // Current delivered by generator 1 in A\n",
+"I_L2 = (E_g2 - V_L)/R_a2 ; // Current delivered by generator 2 in A\n",
+"I_L3 = (E_g3 - V_L)/R_a3 ; // Current delivered by generator 3 in A\n",
+"I_L_2ohm = V_L / R_L ; // Current delivered by 2 ohm load in A\n",
+"\n",
+"// case c\n",
+"I_a1 = I_L1 ; // Armature current in A for generator 1\n",
+"I_a2 = I_L2 ; // Armature current in A for generator 2\n",
+"I_a3 = I_L3 ; // Armature current in A for generator 3\n",
+"\n",
+"P_g1 = E_g1 * I_a1 ; // Power generated by generator 1 in W\n",
+"P_g2 = E_g2 * I_a2 ; // Power generated by generator 2 in W\n",
+"P_g3 = E_g3 * I_a3 ; // Power generated by generator 3 in W\n",
+"\n",
+"// case d\n",
+"P_L1 = V_L * I_L1 ; // Power delivered to or received from generator 1 in W\n",
+"P_L2 = V_L * I_L2 ; // Power delivered to or received from generator 2 in W\n",
+"P_L3 = V_L * I_L3 ; // Power delivered to or received from generator 3 in W\n",
+"P_L = V_L * -I_L_2ohm ; // Power delivered to or received 2 ohm load in W\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-3 Solution : ');\n",
+"printf(' \n a: Converting each voltage source to a current source and applying');\n",
+"printf(' \n    Millman`s theorem yields ')\n",
+"printf(' \n    V_L = %d V \n ', V_L );\n",
+"\n",
+"printf(' \n b: I_L1 = %d A (to bus)', I_L1 );\n",
+"printf(' \n    I_L2 = %d A ', I_L2 );\n",
+"printf(' \n    I_L3 = %d A (from bus)', I_L3 );\n",
+"printf(' \n    I_L_2ohm = -%d A (from bus) \n', I_L_2ohm );\n",
+"\n",
+"printf(' \n c: P_g1 = %d W ',P_g1 );\n",
+"printf(' \n    P_g2 = %d W (floating)',P_g2 );\n",
+"printf(' \n    P_g3 = %d W \n',P_g3 );\n",
+"\n",
+"printf(' \n d: P_L1 = %d W ', P_L1);\n",
+"printf(' \n    P_L2 = %d W ', P_L2 );\n",
+"printf(' \n    P_L3 = %d W ', P_L3 );\n",
+"printf(' \n    P_L = %d W ', P_L );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.4: calculate_total_load_and_kW_output_of_each_G.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P1 = 300 ; // Power rating of generator 1 in kW\n",
+"P2 = 600 ; // Power rating of generator 2 in kW\n",
+"V = 220 ; // Voltage rating of generator 1 and 2 in volt\n",
+"V_o = 250 ; // No-load voltage applied to both the generators in volt\n",
+"// Assume linear characteristics\n",
+"V_1 = 230 ; // Terminal voltage in volt (case a)\n",
+"V_2 = 240 ; // Terminal voltage in volt (case b)\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"kW1_a = (V_o - V_1)/(V_o - V) * P1 ; // kW carried by generator 1\n",
+"kW2_a = (V_o - V_1)/(V_o - V) * P2 ; // kW carried by generator 2\n",
+"\n",
+"// case b\n",
+"kW1_b = (V_o - V_2)/(V_o - V) * P1 ; // kW carried by generator 1\n",
+"kW2_b = (V_o - V_2)/(V_o - V) * P2 ; // kW carried by generator 2\n",
+"\n",
+"// case c\n",
+"frac_a = (V_o - V_1)/(V_o - V); // Fraction of rated kW carried by each generator\n",
+"frac_b = (V_o - V_2)/(V_o - V); // Fraction of rated kW carried by each generator\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-4 Solution : ');\n",
+"printf(' \n a: At 230 V, using Eq.(7-3) below : ');\n",
+"printf(' \n    Generator 1 carries = %d kW ', kW1_a );\n",
+"printf(' \n    Generator 2 carries = %d kW \n', kW2_a );\n",
+"\n",
+"printf(' \n b: At 240 V, using Eq.(7-3) below : ');\n",
+"printf(' \n    Generator 1 carries = %d kW ', kW1_b );\n",
+"printf(' \n    Generator 2 carries = %d kW \n', kW2_b );\n",
+"\n",
+"printf(' \n c: Both generators carry no-load at 250 V; ');\n",
+"printf(' \n    %f rated load at %d V; ', frac_b , V_2 );\n",
+"printf(' \n    %f rated load at %d V; ', frac_a , V_1 );\n",
+"printf(' \n    and rated load at %d V. ', V );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.5: calculate_max_and_min_E_and_frequency_and_Epeak_and_n.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"E_1 = 220 ; // Terminal voltage of alternator 1 in volt \n",
+"E_2 = 222 ; // Terminal voltage of alternator 2 in volt \n",
+"f_1 = 60 ; // Frequency of alternator 1 in Hz\n",
+"f_2 = 59.5 ; // Frequency of alternator 2 in Hz\n",
+"// Switch is open\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"E_max = (E_1 + E_2)/2 ; // Maximum effective voltage across each lamp in volt \n",
+"E_min = (E_2 - E_1)/2 ; // Minimum effective voltage across each lamp in volt \n",
+"\n",
+"// case b\n",
+"f = f_1 - f_2 ; // Frequency in Hz of the voltage across the lamps\n",
+"\n",
+"// case c\n",
+"E_peak = E_max / 0.7071 ; // Peak value of the voltage in volt across each lamp\n",
+"\n",
+"// case d\n",
+"n = (1/2)*f_1 ; // Number of maximum light pulsations per minute\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-5 Solution : ');\n",
+"printf(' \n a: E_max/lamp = %d V (rms)\n ', E_max );\n",
+"printf(' \n    E_min/lamp = %d V \n ', E_min );\n",
+"printf(' \n b: f = %.1f Hz \n ', f );\n",
+"printf(' \n c: E_peak = %.f V \n ', E_peak );\n",
+"printf(' \n d: n = %d pulsations/min ', n );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.6: calculate_max_and_min_E_and_f_and_phase_relations.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"E = 220 ; // Voltage generated in volt\n",
+"E_1 = E ; // Voltage generated by alternator 1 in volt\n",
+"E_2 = E ; // Voltage generated by alternator 2 in volt\n",
+"f_1 = 60 ; // Frequency in Hz of alternator 1\n",
+"f_2 = 58 ; // Frequency in Hz of alternator 2\n",
+"// Switch is open\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"E_max = (E_1 + E_2)/2 ; // Maximum effective voltage across each lamp in volt\n",
+"f = f_1 - f_2 ; // Frequency in Hz of the voltage across the lamps\n",
+"\n",
+"// case c\n",
+"E_min = (E_2 - E_1)/2 ; // Minimum effective voltage across each lamp in volt \n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-6 Solution : ');\n",
+"printf(' \n a: E_max/lamp = %d V \n    f = %d Hz \n ', E_max, f );\n",
+"printf(' \n b: The voltages are equal and opposite in the local circuit. \n ');\n",
+"printf(' \n c: E_min/lamp = %d V at zero frequency \n ', E_min );\n",
+"printf(' \n d: The voltages are in phase in the local circuit. ');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.7: calculate_Is_in_both_alternators.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data as per Ex.(7-5)\n",
+"E1 = 220 ; // Terminal voltage of alternator 1 in volt \n",
+"E2 = 222 ; // Terminal voltage of alternator 2 in volt \n",
+"f1 = 60 ; // Frequency of alternator 1 in Hz\n",
+"f2 = 59.5 ; // Frequency of alternator 2 in Hz\n",
+"// Switch is open\n",
+"\n",
+"// Given data as per Ex.(7-6)\n",
+"E = 220 ; // Voltage generated in volt\n",
+"E_1 = E ; // Voltage generated by alternator 1 in volt\n",
+"E_2 = E ; // Voltage generated by alternator 2 in volt\n",
+"f_1 = 60 ; // Frequency in Hz of alternator 1\n",
+"f_2 = 58 ; // Frequency in Hz of alternator 2\n",
+"// Switch is open\n",
+"\n",
+"// Given data as per Ex.(7-7)\n",
+"R_a1 = 0.1 ; // armature resistance of alternator 1 in ohm\n",
+"R_a2 = 0.1 ; // armature resistance of alternator 2 in ohm\n",
+"X_a1 = 0.9 ; // armature reactance of alternator 1 in ohm\n",
+"X_a2 = 0.9 ; // armature reactance of alternator 2 in ohm\n",
+"\n",
+"Z_1  = R_a1 + %i*X_a1 ; // Effective impedance of alternator 1 in ohm\n",
+"Z_2  = R_a1 + %i*X_a2 ; // Effective impedance of alternator 2 in ohm\n",
+"// Switches are closed at the proper instant for paralleling.\n",
+"\n",
+"// Calculations\n",
+"// In Ex.7-5,\n",
+"E_r = E2 - E1 ; // Effective voltage generated in volt \n",
+"I_s = E_r / (Z_1 + Z_2); // Synchronizing current in the armature in A\n",
+"I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A\n",
+"I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees\n",
+"\n",
+"// In Ex.7-6,\n",
+"Er = E_2 -E_1 ; // Effective voltage generated in volt \n",
+"Is = Er / ( Z_1 + Z_2); // Synchronizing current in the armature in A\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-7 Solution : ');\n",
+"printf(' \n In Ex.7-5, ');\n",
+"printf(' \n E_r = %d V ', E_r);\n",
+"printf(' \n I_s = ');disp(I_s);\n",
+"printf(' \n I_s = %.3f <%.2f A ',I_s_m, I_s_a);\n",
+"printf(' \n where %.3f is magnitude in A and %.2f is phase angle in degrees \n',I_s_m,I_s_a);\n",
+"\n",
+"printf(' \n In Ex.7-6, ');\n",
+"printf(' \n E_r = %d V ', Er );\n",
+"printf(' \n I_s = %d A',Is);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.8: EX7_8.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// EMF's are opposed exactly by 180 degrees\n",
+"E_gp1 = 200 ; // Terminal voltage of alternator 1 in volt \n",
+"E_gp2 = 220 ; // Terminal voltage of alternator 2 in volt \n",
+"R_a1 = 0.2 ; // armature resistance of alternator 1 in ohm\n",
+"R_a2 = 0.2 ; // armature resistance of alternator 2 in ohm\n",
+"X_a1 = 2 ; // armature reactance of alternator 1 in ohm\n",
+"X_a2 = 2 ; // armature reactance of alternator 2 in ohm\n",
+"\n",
+"Z_p1  = R_a1 + %i*X_a1 ; // Effective impedance of alternator 1 in ohm\n",
+"Z_p2  = R_a1 + %i*X_a2 ; // Effective impedance of alternator 2 in ohm\n",
+"// Switches are closed at the proper instant for paralleling.\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"E_r = (E_gp2 - E_gp1) ; // Effective voltage generated in volt \n",
+"I_s = E_r / (Z_p1 + Z_p2); // Synchronizing current in the armature in A\n",
+"I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A\n",
+"I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees\n",
+"\n",
+"P_2 = E_gp2 * I_s_m * cosd(I_s_a); // Generator action developed by alternator 2 in W\n",
+"\n",
+"// case b\n",
+"theta = I_s_a;\n",
+"// P_1 = E_gp1 * I_s_m * cosd(180 - theta)\n",
+"// P_1 = -E_gp1 * I_s_m * cosd(theta), \n",
+"P_1 = -E_gp1 * I_s_m * cosd(theta); // Synchronizing power received by alternator 1 in W \n",
+"\n",
+"// case c\n",
+"// but consider +ve vlaue for P_1 for finding losses, so\n",
+"P1 = abs(P_1);\n",
+"losses = P_2 - P1 ; // Power losses in both armatures in W\n",
+"check = E_r * I_s_m * cosd(I_s_a); // Verifying losses by Eq.7-7\n",
+"double_check = (I_s_m)^2 * (R_a1 + R_a2); // Verifying losses by Eq.7-7\n",
+"\n",
+"// case d\n",
+"V_p2  = E_gp2 - I_s*Z_p1 ; // Generator action\n",
+"V_p1  = E_gp1 + I_s*Z_p1 ; // Motor action\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-8 Solution : ');\n",
+"printf(' \n a: E_r = %d V ',E_r);\n",
+"printf(' \n    I_s = %.2f <%.2f A ', I_s_m, I_s_a );\n",
+"printf(' \n    P_2 = %.1f W (total power delivered by alternator 2 ) \n', P_2);\n",
+"\n",
+"printf(' \n b: P_1 = %f W (synchronizing power received by alternator 1)',P_1);\n",
+"printf(' \n    Note:Scilab considers phase angle of I_s as %f instead ',I_s_a);\n",
+"printf(' \n         of -84.3 degrees,so slight variation in the answer P_1.\n');\n",
+"\n",
+"printf(' \n c: Consider +ve value of P_1 for calculating losses'); \n",
+"printf(' \n    Losses: P_2 - P_1 = %.1f W ',losses );\n",
+"printf(' \n    Check: E_a*I_s*cos(theta) = %.1f W ',check );\n",
+"printf(' \n    Double check : (I_s)^2*(R_a1+R_a2) = %.1f W as given in Eq.(7-1)',double_check );\n",
+"\n",
+"printf('\n\n d: From Fig.7-14, V_p2, the terminal phase voltage of ');\n",
+"printf(' \n    alternator 2, is, from Eq.(7-1)');\n",
+"printf(' \n    V_p2 = %d V (generator action)\n\n    From section 7-2.1 ',V_p2);\n",
+"printf(' \n    V_p1 = %d V ( motor action)\n',V_p1);\n",
+"\n",
+"printf(' \n e: The phasor diagram is shown in Fig.7-14.');\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 7.9: calculate_synchronizing_I_and_P_and_P_losses.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 7: PARALLEL OPERATION\n",
+"// Example 7-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"E_2_mag = 230 ; // Magnitude of voltage generated by alternator 2 in volt\n",
+"E_1_mag = 230 ; // Magnitude of voltage generated by alternator 1 in volt\n",
+"\n",
+"theta_2 = 180 ; // Phase angle of generated voltage by alternator 2 in degrees\n",
+"theta_1 = 20 ; // Phase angle of generated voltage by alternator 1 in degrees\n",
+"\n",
+"R_a1 = 0.2 ; // armature resistance of alternator 1 in ohm\n",
+"R_a2 = 0.2 ; // armature resistance of alternator 2 in ohm\n",
+"\n",
+"// writing given voltage in exponential form as follows\n",
+"// %pi/180 for degrees to radians conversion\n",
+"E_2 = E_2_mag * expm(%i * theta_2*(%pi/180) ); // voltage generated by alternator 2 in volt\n",
+"E_1 = E_1_mag * expm(%i * theta_1*(%pi/180) ); // voltage generated by alternator 1 in volt\n",
+"\n",
+"// writing given impedance(in ohm)in exponential form as follows\n",
+"Z_1 = 2.01 * expm(%i * 84.3*(%pi/180) ); // %pi/180 for degrees to radians conversion\n",
+"Z_2 = Z_1 ;\n",
+"Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees\n",
+"\n",
+"// Calculations\n",
+"E_r = E_2 + E_1 ; // Total voltage generated by Alternator 1 and 2 in volt\n",
+"E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt\n",
+"E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees\n",
+"\n",
+"// case a\n",
+"I_s = E_r / (Z_1 + Z_2); // Synchronozing current in A\n",
+"I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A\n",
+"I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees\n",
+"\n",
+"// case b\n",
+"E_gp1 = E_1_mag;\n",
+"P_1 = E_gp1 * I_s_m * cosd(I_s_a - theta_1); // Synchronozing power developed by alternator 1 in W\n",
+"\n",
+"// case c\n",
+"E_gp2 = E_2_mag;\n",
+"P_2 = E_gp2 * I_s_m * cosd(I_s_a - theta_2); // Synchronozing power developed by alternator 2 in W\n",
+"\n",
+"// case d\n",
+"// but consider +ve vlaue for P_2 for finding losses, so\n",
+"P2 = abs(P_2);\n",
+"losses = P_1 - P2 ; // Losses in the armature in W\n",
+"\n",
+"// E_r_a yields -80 degrees which is equivalent to 100 degrees, so\n",
+"theta = 100 - I_s_a ; // Phase difference between E_r and I_a in degrees\n",
+"\n",
+"check = E_r_m * I_s_m * cosd(theta); // Verifying losses by Eq.7-7\n",
+"R_aT = R_a1 + R_a2 ; // total armature resistance of alternator 1 and 2 in ohm\n",
+"double_check = (I_s_m)^2 * (R_aT); // Verifying losses by Eq.7-7\n",
+"\n",
+"// Display the results\n",
+"disp('Example 7-9 Solution : ');\n",
+"printf(' \n a: I_s = ');disp(I_s);\n",
+"printf(' \n    I_s = %.2f <%.2f A \n ',I_s_m, I_s_a );\n",
+"\n",
+"printf(' \n b: P_1 = %.f W (power delivered to bus)',P_1);\n",
+"printf(' \n    Slight variation in P_1 is due slight variations in ')\n",
+"printf(' \n    magnitude of I_s,& angle btw (E_gp1,I_s)\n')\n",
+"printf(' \n    P_2 = %.f W (power received from bus)\n',P_2);\n",
+"\n",
+"printf(' \n c: Losses: P_1 - P_2 = %d',losses);\n",
+"printf(' \n    Check: E_a*I_s*cos(theta) = %d W ',check );\n",
+"printf(' \n    Double check : (I_s)^2*(R_a1+R_a2) = %d W ',double_check );"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/8-AC_DYNAMO_TORQUE_RELATIONS_SYNCHRONOUS_MOTORS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/8-AC_DYNAMO_TORQUE_RELATIONS_SYNCHRONOUS_MOTORS.ipynb
new file mode 100644
index 0000000..ff9d0d2
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/8-AC_DYNAMO_TORQUE_RELATIONS_SYNCHRONOUS_MOTORS.ipynb
@@ -0,0 +1,1710 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 8: AC DYNAMO TORQUE RELATIONS SYNCHRONOUS MOTORS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.10: calculate_cost_of_raising_PF_to_unity_and_point85_lagging.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-10\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"kVA = 10000 ; // kVA rating of a system\n",
+"cos_theta = 0.65 ; // power factor of the system\n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"cos_theta_b = 0.85 ; // Raised PF\n",
+"sin_theta_b = sqrt( 1 - (cos_theta_b)^2 );\n",
+"cost = 60 ; // cost of the synchronous capacitor to improve the PF in dollars/kVA\n",
+"// neglect the losses in the synchronous capacitor\n",
+"// Calculations\n",
+"// case a : For unity PF\n",
+"// at the original load\n",
+"kW_a = kVA * cos_theta ; //\n",
+"theta = acosd(cos_theta) ; // Power factor angle of the system in degrees\n",
+"kvar = kVA * sind(theta) ; // Reactive power in kvar\n",
+"kVA_a = kvar ;\n",
+"cost_cap_a = kvar * cost ; // Cost of raising the PF to unity PF in dollars\n",
+"// case b\n",
+"theta_b = acosd(cos_theta_b) ; // Power factor angle of the system in degrees\n",
+"kVA_b = kW_a / cos_theta_b ; // kVA value reduction\n",
+"kvar_b = kVA_b * sind(theta_b) ; // final kvar value reduced\n",
+"kvar_add = kvar - kvar_b ; // kvar of correction added\n",
+"cost_cap_b = kvar_add * cost ; // Cost of raising the PF to 0.85 PF in dollars\n",
+"// Display the results\n",
+"disp('Example 8-10 Solution : ');\n",
+"printf(' \n    Note : Slight variations in the kvar and cost values are due to  ');\n",
+"printf(' \n    non-approximation of theta values while calculating in scilab.\n');\n",
+"printf(' \n a: At the original load,\n');\n",
+"printf(' \n    kW = %d kW at theta = %.1f degrees \n', kW_a , theta );\n",
+"printf(' \n    kvar = %.3f kvar\n\n    For unity PF,',kvar);\n",
+"printf(' \n    kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n',kVA_a);\n",
+"printf(' \n    Cost of synchronous capacitor = $%.f \n\n',cost_cap_a );\n",
+"printf(' \n b: For %.2f, PF = cos(%.1f), the total power,',cos_theta_b, theta_b);\n",
+"printf(' \n    %.f kW,remains the same. Therefore,\n ',kW_a);\n",
+"printf(' \n    kVA of final system reduced to = %.f kVA \n',kVA_b);\n",
+"printf(' \n    kvar of final system reduced to = %.f kvar \n    Therefore,',kvar_b);\n",
+"printf(' \n    kvar of correction added = %.3f kvar\n ',kvar_add);\n",
+"printf(' \n    kVA of synchronous capacitor = %.3f kVA (neglecting losses)\n',kvar_add);\n",
+"printf(' \n    Cost of synchronous capacitor = $%.f',cost_cap_b );\n",
+"printf(' \n    or less than half the cost in part(a)');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.11: calculate_Po_jQo_and_power_triangle.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"S_conjugate = 1000 ; // Apparent complex power in kVA\n",
+"cos_theta = 0.6 ; // lagging PF\n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"P_o = S_conjugate * cos_theta ; // Active power dissipated by the load in kW\n",
+"\n",
+"// case b\n",
+"jQ_o = S_conjugate * sin_theta ; // Inductive reactive quadrature power - \n",
+"// - drawn from and returned to the supply\n",
+"\n",
+"// Display the results\n",
+"\n",
+"disp('Example 8-11 Solution : ');\n",
+"printf(' \n a: Active power \n    P_o = %d kW \n ', P_o );\n",
+"\n",
+"printf(' \n b: Inductive reactive quadrature power \n    +jQ_o in kvar =  \n');disp(%i*jQ_o);\n",
+"\n",
+"printf(' \n c: The original power triangle is shown in Fig.8-26a.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.12: calculate_Pf_jQf_Pa_jQa_kVA_and_draw_power_tabulation_grid.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-12\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"S_conjugate = 1000 ; // Apparent complex power in kVA\n",
+"cos_theta_f = 0.8 ; // lagging PF\n",
+"sin_theta_f = sqrt( 1 - (cos_theta_f)^2 );\n",
+"\n",
+"// Calculated values from Ex.8-11\n",
+"P_o = 600 ; // Active power dissipated by the load in kW\n",
+"Q_o = 800 ; // Inductive reactive quadrature power - \n",
+"// - drawn from and returned to the supply \n",
+"\n",
+"// Calculations :\n",
+" \n",
+"// case a\n",
+"P_f = S_conjugate * cos_theta_f ; // Active power dissipated by the load in kW\n",
+"\n",
+"// case b\n",
+"Q_f = S_conjugate * sin_theta_f ; // Reactive quadrature power drawn from -\n",
+"//  - and returned to the supply\n",
+"\n",
+"// case c\n",
+"P_a = P_f - P_o ; // Additional active power in kW that may be supplied to -\n",
+"// - new customers\n",
+"\n",
+"// case d\n",
+"jQ_a = %i * ( Q_f ) - %i * ( Q_o ); // Correction kvar required to raise PF -\n",
+"// -from 0.6 to o.8 lagging\n",
+"\n",
+"// case e\n",
+"S_c_conjugate = 0 - jQ_a ; // Rating of correction capacitors needed for case d\n",
+"\n",
+"// Display the results\n",
+"\n",
+"disp('Example 8-12 Solution : ');\n",
+"printf(' \n a: P_f = %d kW \n ', P_f );\n",
+"printf(' \n b: +jQ_f in kvar = ');disp(%i*Q_f);\n",
+"printf(' \n c: P_a = %d kW \n ', P_a );\n",
+"printf(' \n d: jQ_a in kvar =   ');disp(jQ_a)\n",
+"printf(' \n e: S_c_conjugate = %d kVA \n ', abs(S_c_conjugate) );\n",
+"printf(' \n f: The power tabulation grid is shown in Fig.8-26b.');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.13: calculate_Pf_jQf_Pa_jQa_kVA_and_power_tabulation_grid.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-13\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Ex.8-12 PF\n",
+"cos_theta = 0.6 ; // PF lagging\n",
+"\n",
+"// Given data\n",
+"S_conjugate = 1000 ; // Apparent complex power in kVA\n",
+"cos_theta_f = 1.0 ; // unity PF\n",
+"sin_theta_f = sqrt( 1 - (cos_theta_f)^2 );\n",
+"\n",
+"// Calculated values from Ex.8-11\n",
+"P_o = 600 ; // Active power dissipated by the load in kW\n",
+"Q_o = 800 ; // Inductive reactive quadrature power - \n",
+"// - drawn from and returned to the supply \n",
+"\n",
+"// Calculations :\n",
+" \n",
+"// case a\n",
+"P_f = S_conjugate * cos_theta_f ; // Active power dissipated by the load in kW\n",
+"\n",
+"// case b\n",
+"Q_f = S_conjugate * sin_theta_f ; // Reactive quadrature power drawn from -\n",
+"//  - and returned to the supply\n",
+"\n",
+"// case c\n",
+"P_a = P_f - P_o ; // Additional active power in kW that may be supplied to -\n",
+"// - new customers\n",
+"\n",
+"// case d\n",
+"jQ_a = %i * ( Q_f ) - %i * ( Q_o ); // Correction kvar required to raise PF -\n",
+"// -from 0.6 to o.8 lagging\n",
+"Q_a = -abs(jQ_a); // \n",
+"\n",
+"// case e\n",
+"S_c_conjugate = 0 - jQ_a ; // Rating of correction capacitors needed for case d\n",
+"\n",
+"// Display the results\n",
+"\n",
+"disp('Example 8-13 Solution : ');\n",
+"printf(' \n a: P_f = %d kW \n ', P_f );\n",
+"printf(' \n b: +jQ_f in kvar = ');disp(%i*Q_f);\n",
+"printf(' \n c: P_a = %d kW \n ', P_a );\n",
+"printf(' \n d: jQ_a in kvar =   ');disp(jQ_a)\n",
+"printf(' \n e: S_c_conjugate = %d kVA \n ', abs(S_c_conjugate) );\n",
+"printf(' \n f: The power tabulation grid is shown below.\n');\n",
+"printf(' \n    \t\t P \t ±jQ \t S* ');\n",
+"printf(' \n    \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ');\n",
+"printf(' \n    ___________________________________________');\n",
+"printf(' \n    Original : \t %d \t +j%d \t %d \t %.1f ',P_o ,Q_o ,S_conjugate,cos_theta);\n",
+"printf(' \n    Added : \t %d \t %dj \t __ \t __',P_a ,Q_a );\n",
+"printf(' \n    Final : \t %d \t +j%d \t %d \t %.1f',P_f ,Q_f ,S_conjugate,cos_theta_f);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.14: calculate_original_and_final_kVA_kvar_P_and_correction_kvar_Sa.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-14\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 2000 ; // load in kW drawn by a factory\n",
+"cos_theta_o = 0.6 ; // PF lagging\n",
+"sin_theta_o = sqrt( 1-  (cos_theta_o)^2 );\n",
+"cos_theta_f = 0.85 ; // final PF lagging required\n",
+"sin_theta_f = sqrt( 1-  (cos_theta_f)^2 );\n",
+"P_a = 275 ; // Losses in the synchronous capacitor in kW\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"S_o_conjugate = P_o / cos_theta_o ; // Original kVA drawn from the utility \n",
+"\n",
+"// case b\n",
+"Q_o = S_o_conjugate * sin_theta_o ; // Original lagging kvar\n",
+"\n",
+"// case c\n",
+"P_f = P_o + P_a ; // Final system active power consumed from the utility in kW\n",
+"\n",
+"// case d\n",
+"S_f_conjugate = P_f / cos_theta_f ; // Final kVA drawn from the utility \n",
+"S_f_conjugate_a = acosd(cos_theta_f); // Phase angle of S_f_conjugate in degrees \n",
+"\n",
+"// case e\n",
+"jQ_f = S_f_conjugate * sin_theta_f ; // Final lagging kvar\n",
+"jQ_a = %i*(jQ_f) - %i*(Q_o); // Correction kvar produced by the synchronous capacitor\n",
+"Q_a = abs(jQ_a); // Magnitude of jQ_a in kvar \n",
+"\n",
+"// case f\n",
+"P = P_a ;\n",
+"S_a_conjugate = P -%i*(abs(jQ_a)); // kVA rating of the synchronous capacitor\n",
+"S_a_conjugate_m = abs(S_a_conjugate);//S_a_conjugate_m = magnitude of S_a_conjugate in kVA\n",
+"S_a_conjugate_a = atan(imag(S_a_conjugate) /real(S_a_conjugate))*180/%pi;\n",
+"//S_a_conjugate_a=phase angle of S_a_conjugate in degrees\n",
+"PF_f = cosd(S_a_conjugate_a); // PF\n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-14 Solution : ');\n",
+"printf(' \n a: S*o = %.1f kVA \n',S_o_conjugate );\n",
+"\n",
+"printf(' \n b: Q*o in kvar = ' );disp(%i*Q_o);\n",
+"\n",
+"printf(' \n c: P*f = %.f kW \n',P_f );\n",
+"\n",
+"printf(' \n d: S*f = %.1f <%.1f kVA\n ',S_f_conjugate,S_f_conjugate_a );\n",
+"\n",
+"printf(' \n e: jQ_f in kvar = ');disp(%i*jQ_f);\n",
+"printf(' \n    -jQ_a in kvar = ');disp(jQ_a);\n",
+"\n",
+"printf(' \n f: S*a = %.f <%.2f kVA ', S_a_conjugate_m , S_a_conjugate_a );\n",
+"printf(' \n    (cos(%.2f) = %.3f leading)\n',S_a_conjugate_a,PF_f);\n",
+"\n",
+"printf(' \n g: Power tabulation grid : \n ');\n",
+"printf(' \n    \t\t P \t ±jQ \t S* ');\n",
+"printf(' \n    \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ');\n",
+"printf(' \n    ___________________________________________');\n",
+"printf(' \n    Original : \t %d \t +j%.f  %.1f  %.1f lag',P_o ,Q_o ,S_o_conjugate,cos_theta_o);\n",
+"printf(' \n    Added    : \t %d \t -%.fj  %.f \t %.3f lead',P_a ,Q_a,S_a_conjugate_m,cosd(S_a_conjugate_a) );\n",
+"printf(' \n    Final    : \t %d \t +j%.f  %.1f  %.2f lag',P_f ,jQ_f ,S_f_conjugate,cos_theta_f);\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.15: calculate_kVA_added_Pa_and_Qa_and_Pf_Qf_and_PF.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-15\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 2275 ; // Original kVA \n",
+"Q_o = 1410 ; // Original kvar \n",
+"S_f_conjugate = 3333.3 ; // final kVA of the load\n",
+"S_o_conjugate = P_o + %i*Q_o ; // Load of the alternator in kVA\n",
+"S_o_conjugate_m = abs(S_o_conjugate);//S_o_conjugate_m = magnitude of S_o_conjugate in kVA\n",
+"S_o_conjugate_a = atan(imag(S_o_conjugate) /real(S_o_conjugate))*180/%pi;\n",
+"//S_o_conjugate_a=phase angle of S_o_conjugate in degrees\n",
+"\n",
+"disp('Example 8-15');\n",
+"printf(' \n  Power tabulation grid : \n ');\n",
+"printf(' \n  \t\t P \t\t ±jQ \t\t S* ');\n",
+"printf(' \n  \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ');\n",
+"printf(' \n  ________________________________________________________________________');\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));\n",
+"printf(' \n  Added : \t0.8x \t\t j0.6x \t\t x \t\t0.80 lag' );\n",
+"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 );\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// Assume x is the additional kVA load. Then real and quadrature powers are 0.8x and j0.6x\n",
+"// respectively,as shown. Adding each column vertically and using the Pythagorean theorem,\n",
+"// we may write (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this eqution yields\n",
+"// the quadratic x^2 + 5352x -3947163 = 0. Applying the quadratic yields the added kVA load:\n",
+"x = poly(0,'x'); // Defining a polynomial with variable 'x' with  root at 0\n",
+"p = -3947163 + 5352*x + x^2 \n",
+"a = 1 ; // coefficient of x^2\n",
+"b = 5332 ; // coefficient of x\n",
+"c = -3947163 ; // constant\n",
+"\n",
+"// Roots of p\n",
+"x1 = ( -b + sqrt (b^2 -4*a*c ) ) /(2* a);\n",
+"x2=( -b - sqrt (b^2 -4*a*c ) ) /(2* a);\n",
+"\n",
+"// case b\n",
+"P_a = 0.8*x1 ; // Added active power of the additional load in kW\n",
+"Q_a = 0.6*x1 ; // Added reactive power of the additional load in kvar\n",
+"\n",
+"// case c\n",
+"P_f = P_o + P_a ; // Final active power of the additional load in kW\n",
+"Q_f = Q_o + Q_a ; // Final reactive power of the additional load in kvar \n",
+"\n",
+"// case d\n",
+"PF = P_f / S_f_conjugate ; // Final power factor\n",
+"// Validity check\n",
+"S_conjugate_f = P_f + %i*Q_f ; // Final kVA of the load \n",
+"S_conjugate_f_m = abs(S_conjugate_f);//S_conjugate_f_m = magnitude of S_conjugate_f in kVA\n",
+"S_conjugate_f_a = atan(imag(S_conjugate_f) /real(S_conjugate_f))*180/%pi;\n",
+"//S_conjugate_f_a=phase angle of S_conjugate_f in degrees\n",
+"\n",
+"// Display the results\n",
+"\n",
+"disp(' Solution : ')\n",
+"\n",
+"printf(' \n a: The given data is shown in the above power tabulation grid.Assume');\n",
+"printf(' \n    x is the additional kVA load. Then real and quadrature powers are');\n",
+"printf(' \n    0.8x and j0.6x respectively,as shown.Adding each column vertically');\n",
+"printf(' \n    and using the Pythagorean theorem, we may write');\n",
+"printf(' \n    (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this');\n",
+"printf(' \n    equation yields the quadratic as follows : \n');\n",
+"printf(' \n    x^2 + 5332x -3947163 = 0. \n ')\n",
+"printf(' \n    Applying the quadratic yields the added kVA load:');\n",
+"printf(' \n    Roots of quadratic Eqn p are \n ');\n",
+"printf(' \n    x1 = %.2f \n    x2 = %.2f ', x1, x2 );\n",
+"printf(' \n    Consider +ve value of x for added kVA so');\n",
+"printf(' \n    x = S*a = %.2f kVA \n ', x1 );\n",
+"\n",
+"printf(' \n b: P_a = %.1f kW \n ', P_a );\n",
+"printf(' \n    Q_a in kvar = \n');disp(%i*Q_a);\n",
+"\n",
+"printf(' \n c: P_f = %.1f kW \n ', P_f );\n",
+"printf(' \n    Q_f in kvar = \n');disp(%i*Q_f);\n",
+"\n",
+"printf(' \n d: PF = cosθ_f = %.3f lagging \n ', PF );\n",
+"printf(' \n    Validity check\n    S*f = ');disp(S_conjugate_f);\n",
+"printf(' \n    S*f = %.1f <%.2f kVA \n',S_conjugate_f_m,S_conjugate_f_a);\n",
+"printf(' \n    PF = cos(%.1f) = %.3f lagging',S_conjugate_f_a ,cosd(S_conjugate_f_a));"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.16: Verify_tellegens_theorem_for_kVAs_found_in_Ex_8_15.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-16\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Calculated values as per Ex.8-15 are as follows\n",
+"S_o_conjugate = 2676.5*exp(%i*31.79*(%pi/180)); // Original kVA rating\n",
+"S_o_conjugate_m = abs(S_o_conjugate);//S_o_conjugate_m = magnitude of S_o_conjugate in kVA\n",
+"S_o_conjugate_a = atan(imag(S_o_conjugate) /real(S_o_conjugate))*180/%pi;\n",
+"//S_o_conjugate_a=phase angle of S_o_conjugate in degrees\n",
+"\n",
+"S_a_conjugate = 658.86*exp(%i*36.87*(%pi/180)); // Added kVA rating\n",
+"S_a_conjugate_m = abs(S_a_conjugate);//S_a_conjugate_m = magnitude of S_a_conjugate in kVA\n",
+"S_a_conjugate_a = atan(imag(S_a_conjugate) /real(S_a_conjugate))*180/%pi;\n",
+"//S_a_conjugate_a=phase angle of S_a_conjugate in degrees\n",
+"\n",
+"S_f_conjugate = -3333.3*exp(%i*32.792687*(%pi/180)); // Final kVA rating\n",
+"S_f_conjugate_m = abs(S_f_conjugate);//S_f_conjugate_m = magnitude of S_f_conjugate in kVA\n",
+"S_f_conjugate_a = atan(imag(S_f_conjugate) /real(S_f_conjugate))*180/%pi;\n",
+"//S_f_conjugate_a=phase angle of S_f_conjugate in degrees\n",
+"\n",
+"// Calculations\n",
+"kVA_total = S_o_conjugate + S_a_conjugate + S_f_conjugate ; // Tellegan's theorem\n",
+"kVA_total_m = abs(kVA_total);//kVA_total_m = magnitude of kVA_total in kVA\n",
+"kVA_total_a = atan(imag(kVA_total) /real(kVA_total))*180/%pi;\n",
+"//kVA_total_a=phase angle of kVA_total in degrees\n",
+"\n",
+"// Display the result\n",
+"disp('Example 8-16 Solution : ');\n",
+"printf(' \n From the solution to Ex.8-15, we have ');\n",
+"printf(' \n S*o = %.1f <%.2f kVA \n ', S_o_conjugate_m,S_o_conjugate_a );\n",
+"printf(' \n S*a = %.1f <%.2f kVA \n ', S_a_conjugate_m,S_a_conjugate_a );\n",
+"printf(' \n S*f = %.1f <%.2f kVA \n ', S_f_conjugate_m,S_f_conjugate_a );\n",
+"\n",
+"printf(' \n Validity check ');\n",
+"printf(' \n S*o + S*a + S*f =  ');\n",
+"disp(S_o_conjugate),printf('  +'),disp(S_a_conjugate),printf('  +'),disp(S_f_conjugate);\n",
+"printf(' \n   = %d ',kVA_total );\n",
+"printf(' \n Hence, Tellegen`s theorem is proved');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.17: calculate_overall_PF_using_unity_PF_SM.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-17\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kW = 40000 ; // Load on a factory in kW\n",
+"PF = 0.8 ; // power factor lagging of the load\n",
+"cos_theta = PF; \n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"hp = 7500 ; // power rating of the induction motor in hp\n",
+"PF_IM = 0.75 ; // power factor lagging of the induction motor\n",
+"cos_theta_IM = PF_IM; \n",
+"sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );\n",
+"eta = 91*(1/100) ; // Efficiency of IM\n",
+"PF_SM = 1 ; // power factor  of the synchronous motor\n",
+"\n",
+"// Calculations\n",
+"kVA_original = kW / PF ; // Original kVA \n",
+"kvar_original = kVA_original * sin_theta ; // Original kvar\n",
+"\n",
+"kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW\n",
+"kVA_IM = kW_IM / PF_IM ; // Induction motor kVA\n",
+"kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar\n",
+"\n",
+"kvar_final = kvar_original - kvar_IM ; // final kvar\n",
+"kVA_final = kW + %i*(abs(kvar_final)); // final kVA\n",
+"kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA\n",
+"kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;\n",
+"//kVA_final_a=phase angle of kVA_final in degrees\n",
+"\n",
+"PF_final = cosd(kVA_final_a); // Final power factor\n",
+"\n",
+"// Display the result\n",
+"disp('Example 8-17 Solution : ');\n",
+"printf(' \n The synchronous motor operates at the same efficiency as the IM');\n",
+"printf(' \n that has been replaced, and therefore the total power of the system');\n",
+"printf(' \n is unchanged. The solution involves construction of table that shows ')\n",
+"printf(' \n the original condition of the system, the change, and the final condition.\n');\n",
+"printf(' \n Original kVA = %d kVA \n ', kVA_original );\n",
+"printf(' \n Original kvar = \n' );disp(%i*kvar_original);\n",
+"\n",
+"printf(' \n Induction motor kW = %d kW \n ', kW_IM );\n",
+"printf(' \n Induction motor kVA = %.f kVA \n ', kVA_IM );\n",
+"printf(' \n Induction motor kvar = ');disp(%i*kvar_IM)\n",
+"\n",
+"printf(' \n Final kvar = ');disp(%i*kvar_final);\n",
+"printf(' \n Final kVA = ' );disp(kVA_final);\n",
+"printf(' \n Final kVA = %f <%.2f kVA \n ',kVA_final_m,kVA_final_a);\n",
+"\n",
+"printf(' \n Final PF = %.3f lagging \n ', PF_final );\n",
+"\n",
+"printf(' \n __________________________________________________________________________');\n",
+"printf(' \n Power tabulation grid : \n ');\n",
+"printf(' \n \t\t P \t\t ±jQ \t\t S* ');\n",
+"printf(' \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ');\n",
+"printf(' \n __________________________________________________________________________');\n",
+"printf(' \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag',kW ,kvar_original ,kVA_original,PF);\n",
+"printf(' \n Removed  : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag',kW_IM,kvar_IM,kVA_IM,PF_IM);\n",
+"printf(' \n Added    : \t+%.f  \t\t   0     \t%.1f \t\t 1.0 ',kW_IM,kW_IM);\n",
+"printf(' \n Final    : \t%d \t\tj%.f \t\t%.1f \t %.3f lag',kW ,kvar_final ,kVA_final_m,PF_final);\n",
+"printf(' \n __________________________________________________________________________');\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.18: calculate_overall_PF_using_point8_PF_leading_SM.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-18\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kW = 40000 ; // Load on a factory in kW\n",
+"PF = 0.8 ; // power factor lagging of the load\n",
+"cos_theta = PF; \n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"\n",
+"PF_SM = 0.8 ; // power factor leading of the synchronous motor\n",
+"cos_theta_SM = PF_SM; \n",
+"sin_theta_SM = sqrt( 1 - (cos_theta_SM)^2 );\n",
+"hp = 7500 ; // power rating of the induction motor in hp\n",
+"\n",
+"PF_IM = 0.75 ; // power factor lagging of the induction motor\n",
+"cos_theta_IM = PF_IM; \n",
+"sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );\n",
+"\n",
+"eta = 91*(1/100) ; // Efficiency of IM\n",
+"\n",
+"// Calculations\n",
+"kVA_original = kW / PF ; // Original kVA \n",
+"kvar_original = kVA_original * sin_theta ; // Original kvar\n",
+"\n",
+"\n",
+"kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW\n",
+"kVA_IM = kW_IM / PF_IM ; // Induction motor kVA\n",
+"kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar\n",
+"\n",
+"// case a\n",
+"kW_SM = ( hp * 746 ) / ( 1000 * eta ) ; // Synchronous motor kW\n",
+"kVA_SM = kW_SM / PF_SM ; // Synchronous motor kVA\n",
+"kvar_SM = kVA_SM * sin_theta_SM ; // Synchronous motor kvar\n",
+"\n",
+"kvar_final = kvar_original - kvar_IM - kvar_SM ; // final kvar\n",
+"kVA_final = kW + %i*(abs(kvar_final)); // final kVA\n",
+"kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA\n",
+"kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;\n",
+"//kVA_final_a=phase angle of kVA_final in degrees\n",
+"\n",
+"PF_final = cosd(kVA_final_a); // Final power factor\n",
+"\n",
+"// Display the result\n",
+"disp('Example 8-18 Solution : ');\n",
+"\n",
+"printf(' \n Original kVA = %d kVA \n ', kVA_original );\n",
+"printf(' \n Original kvar = \n' );disp(%i*kvar_original);\n",
+"printf(' \n a:');\n",
+"printf(' \n Synchronous motor kW = %d kW \n ', kW_SM );\n",
+"printf(' \n Synchronous motor kVA = %.f kVA \n ', kVA_SM );\n",
+"printf(' \n Synchronous motor kvar = ');disp(-%i*kvar_SM)\n",
+"\n",
+"printf(' \n Final kvar = ');disp(%i*kvar_final);\n",
+"printf(' \n Final kVA = ' );disp(kVA_final);\n",
+"printf(' \n Final kVA = %f <%.2f kVA \n ',kVA_final_m,kVA_final_a);\n",
+"\n",
+"printf(' \n Final PF = %.3f lagging \n ', PF_final );\n",
+"\n",
+"printf(' \n __________________________________________________________________________');\n",
+"printf(' \n Power tabulation grid : \n ');\n",
+"printf(' \n \t\t P \t\t ±jQ \t\t S* ');\n",
+"printf(' \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ');\n",
+"printf(' \n __________________________________________________________________________');\n",
+"printf(' \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag',kW ,kvar_original ,kVA_original,PF);\n",
+"printf(' \n Removed  : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag',kW_IM,kvar_IM,kVA_IM,PF_IM);\n",
+"printf(' \n Added    : \t+%.f  \t\t-j%.2f     \t%.1f \t\t %.1f lead',kW_SM,abs(kvar_SM),kVA_SM,PF_SM);\n",
+"printf(' \n Final    : \t%d \t\tj%.2f \t%.1f \t %.3f lag',kW ,kvar_final ,kVA_final_m,PF_final);\n",
+"printf(' \n __________________________________________________________________________\n\n');\n",
+"\n",
+"printf(' \n b: ');\n",
+"printf(' \n In Ex.8-17, a 6148 kVA, unity PF, 7500 hp synchronous motor is needed.');\n",
+"printf(' \n In Ex.8-18, a 7685 kVA, 0.8 PF leading, 7500 hp synchronous motor is needed.\n');\n",
+"printf(' \n \t Ex.8-18b shows that a 0.8 PF leading,7500 hp synchronous motor ');\n",
+"printf(' \n must be physically larger than a unity PF,7500 hp synchronous motor ');\n",
+"printf(' \n because of its higher kVA rating.');\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.19: calculate_kVA_and_PF_of_system_and_same_for_SM.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-19\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"kVA_load = 500 ; // Load of 500 kVA\n",
+"PF_load = 0.65 ; // Load operates at this PF lagging\n",
+"cos_theta_load = PF_load ;\n",
+"sin_theta_load = sqrt(1 - (cos_theta_load)^2);\n",
+"hp = 200 ; // power rating of the system in hp\n",
+"eta = 88*(1/100); // Efficiency of the system after adding the load\n",
+"PF_final = 0.85 ; // Final lagging PF after adding the load\n",
+"\n",
+"// Calculations\n",
+"kW_original = kVA_load * cos_theta_load ; // Original kW\n",
+"kvar_original = kVA_load * sin_theta_load ; // Original kvar\n",
+"\n",
+"kW_SM = ( hp * 746 ) / ( 1000 * eta ) ; // Synchronous motor kW\n",
+"\n",
+"// case a\n",
+"kW_final = kW_original + kW_SM ; // final kW of the system with the motor added\n",
+"kVA_final = kW_final / PF_final ; // final kVA of the system with the motor added\n",
+"PF_system = kW_final / kVA_final ; // Final PF of the system with the motor added\n",
+"cos_theta_system = PF_system ; // Final PF of the system with the motor added\n",
+"sin_theta_system  = sqrt(1 - (cos_theta_system )^2);\n",
+"\n",
+"kvar_final = kVA_final * sin_theta_system ; // final kvar of the system with the motor added\n",
+"\n",
+"// case b\n",
+"kvar_SM = %i*kvar_final - %i*kvar_original ; // kvar rating of the sychronous motor\n",
+"\n",
+"kVA_SM = kW_SM + kvar_SM ; // kVA rating of the sychronous motor\n",
+"kVA_SM_m = abs(kVA_SM);//kVA_SM_m = magnitude of kVA_SM in kVA\n",
+"kVA_SM_a = atan(imag(kVA_SM) /real(kVA_SM))*180/%pi;\n",
+"//kVA_SM_a=phase angle of kVA_SM in degrees\n",
+"\n",
+"PF_SM = cosd(kVA_SM_a); // PF of the sychronous motor\n",
+"\n",
+"// Display the result\n",
+"disp('Example 8-19 Solution : ');\n",
+"\n",
+"printf(' \n Original kW = %.f kW \n ', kW_original );\n",
+"printf(' \n Original kvar = %.f kvar\n',kvar_original );\n",
+"printf(' \n Synchronous motor kW = %.1f kW \n ', kW_SM );\n",
+"\n",
+"printf(' \n a: Final kW = %.1f kW',kW_final);\n",
+"printf(' \n    Final kVA of the system = %.f kVA',kVA_final);\n",
+"printf(' \n    System PF = %.2f lagging',PF_system);\n",
+"printf(' \n    Final kvar of the system = j%d (lagging)kvar\n\n',kvar_final);\n",
+"\n",
+"printf(' \n b: Synchronous motor kvar = -%.2fj(leading)kvar\n',abs(kvar_SM));\n",
+"printf(' \n    Synchronous motor kVA = ' );disp(kVA_SM);\n",
+"printf(' \n    Synchronous motor kVA = %.f <%.1f kVA \n ', kVA_SM_m , kVA_SM_a );\n",
+"printf(' \n    Synchronous motor PF = cos(%.1f) = %.3f leading \n ',kVA_SM_a,PF_SM );\n",
+"\n",
+"printf(' \n    ________________________________________________');\n",
+"printf(' \n    Power tabulation grid : \n ');\n",
+"printf(' \n    \t\t P \t ±jQ \t S* ');\n",
+"printf(' \n    \t\t(kW) \t(kvar) \t(kVA) \t cosӨ ');\n",
+"printf(' \n    ________________________________________________');\n",
+"printf(' \n    Original : \t %d \t +j%.f   %.1d \t %.2f lag',kW_original,kvar_original,kVA_load,PF_load);\n",
+"printf(' \n    Added    : \t %.1f \t -%.1fj  %.f \t %.4f lead',kW_SM ,abs(kvar_SM),kVA_SM_m,PF_SM);\n",
+"printf(' \n    Final    : \t %.1f \t +j%.f   %.f     %.2f lag',kW_final,kvar_final,kVA_final,PF_final);\n",
+"printf(' \n    ________________________________________________');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.1: calculate_alpha_Er_Ia_Pp_Pt_Power_loss_Pd.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3- phase Y-connected synchronous motor\n",
+"P = 20 ; // No. of poles\n",
+"hp = 40 ; // power rating of the synchronous motor in hp\n",
+"V_L = 660 ; // Line voltage in volt\n",
+"beta = 0.5 ; // At no-load, the rotor is retarded 0.5 mechanical degree from \n",
+"// its synchronous position.\n",
+"X_s = 10 ; // Synchronous reactance in ohm\n",
+"R_a = 1.0 ; // Effective armature resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"funcprot(0); // To avoid this message 'Warning : redefining function: beta'\n",
+"alpha = P * (beta/2); // The rotor shift from the synchronous position in\n",
+"// electrical degrees.\n",
+"\n",
+"// case b\n",
+"V_p = V_L / sqrt(3); // Phase voltage in volt\n",
+"E_gp = V_p ; // Generated voltage/phase at no-load in volt (given)\n",
+"E_r = (V_p - E_gp*cosd(alpha)) + %i*(E_gp*sind(alpha));\n",
+" // Resultant emf across the armature per phase in V/phase\n",
+"E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt\n",
+"E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees\n",
+"\n",
+"// case c\n",
+"Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm\n",
+"Z_s_m = abs(Z_s);//Z_s_m=magnitude of Z_s in ohm\n",
+"Z_s_a = atan(imag(Z_s) /real(Z_s))*180/%pi;//Z_s_a=phase angle of Z_s in degrees\n",
+"\n",
+"I_a  = E_r / Z_s ; // Armature current/phase in A/phase\n",
+"I_a_m = abs(I_a);//I_a_m=magnitude of I_a in A\n",
+"I_a_a = atan(imag(I_a) /real(I_a))*180/%pi;//I_a_a=phase angle of I_a in degrees \n",
+"\n",
+"// case d\n",
+"theta = I_a_a ; // Phase angle between V_p and I_a in degrees\n",
+"P_p = V_p * I_a_m * cosd(theta); // Power per phase drawn by the motor from the bus\n",
+"P_t = 3*P_p ; // Total power drawn by the motor from the bus\n",
+"\n",
+"// csae e\n",
+"P_a = 3 * (I_a_m)^2 * R_a ; // Armature power loss at no-load in W\n",
+"P_d = (P_t - P_a)/746 ; // Internal developed horsepower at no-load\n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-1 Solution : ');\n",
+"printf(' \n a: alpha = %d degrees (electrical degrees)\n',alpha );\n",
+"\n",
+"printf(' \n b: E_gp = %d V also, as given ',E_gp);\n",
+"printf(' \n    E_r in V/phase = ');disp(E_r);\n",
+"printf(' \n    E_r = %.1f <%.1f V/phase \n',E_r_m, E_r_a );\n",
+"\n",
+"printf(' \n c: Z_s in ohm/phase = ');disp(Z_s);\n",
+"printf(' \n    Z_s = %.2f <%.1f ohm/phase \n',Z_s_m, Z_s_a );\n",
+"printf(' \n    I_a in A/phase = ');disp(I_a);\n",
+"printf(' \n    I_a = %.2f <%.2f A/phase \n ',I_a_m, I_a_a);\n",
+"\n",
+"printf(' \n d: P_p = %.2f W/phase ',P_p );\n",
+"printf(' \n    P_t = %.2f W ',P_t);\n",
+"printf(' \n    Note: Slight variations in power values is due to slight variations');\n",
+"printf(' \n          in V_p , I_a and theta values from those of the textbook\n');\n",
+"\n",
+"printf(' \n e: P_a = %.f W ',P_a );\n",
+"printf(' \n    P_d = %d hp ', P_d );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.20: calulate_speeds_and_poles_for_alternator_and_motor.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-20\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"f_a = 400 ; // Frequency of the alternator in Hz\n",
+"f_m = 60 ; // Frequency of the motor in Hz\n",
+"\n",
+"// Calculations\n",
+"Pole_ratio = f_a / f_m ; // Ratio of no. of poles in alternator to that of motor\n",
+"// Subscript 1 below indicates 1st combination\n",
+"P_a1 = 40 ; // first combination must have 40 poles on the alternator\n",
+"P_m1 = 6 ; // first combination must have 6 poles on the synchronous motor at a speed\n",
+"S_m1 = (120*f_m) / P_m1 ; // Speed of the motor in rpm\n",
+"\n",
+"// Subscript 2 below indicates 2nd combination\n",
+"P_a2 = 80 ; // second combination must have 40 poles on the alternator\n",
+"P_m2 = 12 ; // second combination must have 12 poles on the synchronous motor at a speed\n",
+"S_m2 = (120*f_m) / P_m2 ; // Speed of the motor in rpm\n",
+"\n",
+"// Subscript 13below indicates 3rd combination\n",
+"P_a3 = 120 ; // third combination must have 40 poles on the alternator\n",
+"P_m3 = 18 ; // third combination must have 18 poles on the synchronous motor at a speed\n",
+"S_m3 = (120*f_m) / P_m3 ; // Speed of the motor in rpm\n",
+"\n",
+"// Display the result\n",
+"disp('Example 8-20 Solution : ');\n",
+"\n",
+"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);\n",
+"printf(' \n f_a/f_m determines the combinations of poles and speed.\n');\n",
+"printf(' \n Only even multiples of the above ratio are possible,since poles ');\n",
+"printf(' \n are always in pairs, hence first three combinations are as follows \n');\n",
+"\n",
+"printf(' \n The first combination must have %d poles on the alternator and ',P_a1);\n",
+"printf(' \n %d poles on the sychronous motor at a speed = %d rpm.\n',P_m1,S_m1);\n",
+"\n",
+"printf(' \n The second combination must have %d poles on the alternator and ',P_a2);\n",
+"printf(' \n %d poles on the sychronous motor at a speed = %d rpm.\n',P_m2,S_m2);\n",
+"\n",
+"printf(' \n The third combination must have %d poles on the alternator and ',P_a3);\n",
+"printf(' \n %d poles on the sychronous motor at a speed = %d rpm.\n',P_m3,S_m3);\n",
+"\n",
+"printf(' \n ___________________________________________________________________');\n",
+"printf(' \n Combination \t  Alternator Poles \t  Motor Poles \t Speed (rpm)');\n",
+"printf(' \n             \t         P_a       \t      P_m      \t     S      ');\n",
+"printf(' \n ___________________________________________________________________');\n",
+"printf(' \n First \t\t:\t  %d\t\t       %d \t    %d',P_a1,P_m1,S_m1);\n",
+"printf(' \n Second\t\t:\t  %d\t\t       %d \t    %d',P_a2,P_m2,S_m2);\n",
+"printf(' \n Third \t\t:\t  %d\t\t       %d \t    %d',P_a3,P_m3,S_m3);\n",
+"printf(' \n ___________________________________________________________________');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.2: calculate_alpha_Er_Ia_Pp_Pt_Power_loss_Pd.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3- phase Y-connected synchronous motor\n",
+"P = 20 ; // No. of poles\n",
+"hp = 40 ; // power rating of the synchronous motor in hp\n",
+"V_L = 660 ; // Line voltage in volt\n",
+"beta = 5 ; // At no-load, the rotor is retarded 0.5 mechanical degree from \n",
+"// its synchronous position.\n",
+"X_s = 10 ; // Synchronous reactance in ohm\n",
+"R_a = 1.0 ; // Effective armature resistance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"funcprot(0); // To avoid this message 'Warning : redefining function: beta'\n",
+"alpha = P * (beta/2); // The rotor shift from the synchronous position in\n",
+"// electrical degrees.\n",
+"\n",
+"// case b\n",
+"V_p = V_L / sqrt(3); // Phase voltage in volt\n",
+"E_gp = V_p ; // Generated voltage/phase at no-load in volt (given)\n",
+"E_r = (V_p - E_gp*cosd(alpha)) + %i*(E_gp*sind(alpha));\n",
+"E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt\n",
+"E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees\n",
+"\n",
+"// case c\n",
+"Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm\n",
+"Z_s_m = abs(Z_s);//Z_s_m=magnitude of Z_s in ohm\n",
+"Z_s_a = atan(imag(Z_s) /real(Z_s))*180/%pi;//Z_s_a=phase angle of Z_s in degrees\n",
+"\n",
+"I_a  = E_r / Z_s ; // Armature current/phase in A/phase\n",
+"I_a_m = abs(I_a);//I_a_m=magnitude of I_a in A\n",
+"I_a_a = atan(imag(I_a) /real(I_a))*180/%pi;//I_a_a=phase angle of I_a in degrees \n",
+"\n",
+"// case d\n",
+"theta = I_a_a ; // Phase angle between V_p and I_a in degrees\n",
+"P_p = V_p * I_a_m * cosd(theta); // Power per phase drawn by the motor from the bus\n",
+"P_t = 3*P_p ; // Total power drawn by the motor from the bus\n",
+"\n",
+"// csae e\n",
+"P_a = 3 * (I_a_m)^2 * R_a ; // Armature power loss at no-load in W\n",
+"P_d = (P_t - P_a)/746 ; // Internal developed horsepower at no-load\n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-2 Solution : ');\n",
+"printf(' \n a: alpha = %d degrees (electrical degrees)\n',alpha );\n",
+"\n",
+"printf(' \n b: E_gp = %d V also, as given ',E_gp);\n",
+"printf(' \n    E_r in V/phase = ');disp(E_r);\n",
+"printf(' \n    E_r = %d <%.1f V/phase \n',E_r_m, E_r_a );\n",
+"\n",
+"printf(' \n c: Z_s in ohm/phase = ');disp(Z_s);\n",
+"printf(' \n    Z_s = %.2f <%.1f ohm/phase \n',Z_s_m, Z_s_a );\n",
+"printf(' \n    I_a in A/phase = ');disp(I_a);\n",
+"printf(' \n    I_a = %.2f <%.2f A/phase \n ',I_a_m, I_a_a);\n",
+"\n",
+"printf(' \n d: P_p = %.2f W/phase ',P_p );\n",
+"printf(' \n    P_t = %.2f W ',P_t);\n",
+"printf(' \n    Note: Slight variations in power values is due to slight variations');\n",
+"printf(' \n          in V_p , I_a and theta values from those of the textbook\n');\n",
+"\n",
+"\n",
+"printf(' \n e: P_a = %.f W ',P_a );\n",
+"printf(' \n    P_d = %.1f hp ', P_d );\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.3: calculate_Ia_PF_hp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3- phase Y-connected synchronous motor\n",
+"P = 6 ; // No. of poles\n",
+"hp = 50 ; // power rating of the synchronous motor in hp\n",
+"V_L = 440 ; // Line voltage in volt\n",
+"X_s = 2.4 ; // Synchronous reactance in ohm\n",
+"R_a = 0.1 ; // Effective armature resistance in ohm\n",
+"alpha = 20 ; // The rotor shift from the synchronous position in\n",
+"// electrical degrees.\n",
+"E_gp_a = 240 ; // Generated voltage/phase in volt when the motor is under-excited(case a)\n",
+"E_gp_b = 265 ; // Generated voltage/phase in volt when the motor is under-excited(case b)\n",
+"E_gp_c = 290 ; // Generated voltage/phase in volt when the motor is under-excited(case c)\n",
+"\n",
+"// Calculations\n",
+"V_p = V_L / sqrt(3); // Phase voltage in volt\n",
+"// case a\n",
+"E_ra = (V_p - E_gp_a * cosd(alpha)) + %i*(E_gp_a * sind(alpha));\n",
+"E_ra_m = abs(E_ra);//E_ra_m=magnitude of E_ra in volt\n",
+"E_ra_a = atan(imag(E_ra) /real(E_ra))*180/%pi;//E_ra_a=phase angle of E_ra in degrees\n",
+"\n",
+"Z_s = R_a + %i*X_s ; // Synchronous impedance in ohm\n",
+"\n",
+"I_ap1  = E_ra / Z_s ; // Armature current/phase in A/phase\n",
+"I_ap1_m = abs(I_ap1);//I_ap1_m=magnitude of I_ap1 in A\n",
+"I_ap1_a = atan(imag(I_ap1) /real(I_ap1))*180/%pi;//I_ap1_a=phase angle of I_ap1 in degrees \n",
+"\n",
+"cos_theta_a = cosd(I_ap1_a); // Power factor\n",
+"Ia_m1 = abs(I_ap1_m); // Absoulte value of magnitude of I_ap1\n",
+"\n",
+"P_d1 = 3 * (E_gp_a*Ia_m1) * cosd(160 - I_ap1_a); // // Internal developed power in W\n",
+"// 160 + I_ap1_a is the angle between E_gp_a and I_ap1\n",
+"Pd1 = abs(P_d1); // Consider absolute value of power in W for calculating hp\n",
+"\n",
+"Horse_power1 = Pd1 / 746 ; // Horsepower developed by the armature in hp \n",
+"\n",
+"// case b\n",
+"E_rb = (V_p - E_gp_b * cosd(alpha)) + %i*(E_gp_b * sind(alpha));\n",
+"E_rb_m = abs(E_rb);//E_rb_m=magnitude of E_rb in volt\n",
+"E_rb_a = atan(imag(E_rb) /real(E_rb))*180/%pi;//E_rb_a=phase angle of E_rb in degrees\n",
+"\n",
+"I_ap2  = E_rb / Z_s ; // Armature current/phase in A/phase\n",
+"I_ap2_m = abs(I_ap2);//I_ap2_m=magnitude of I_ap2 in A\n",
+"I_ap2_a = atan(imag(I_ap2) /real(I_ap2))*180/%pi;//I_ap2_a=phase angle of I_ap2 in degrees \n",
+"\n",
+"cos_theta_b = cosd(I_ap2_a); // Power factor\n",
+"Ia_m2 = abs(I_ap2_m); // Absoulte value of magnitude of I_ap2\n",
+"\n",
+"P_d2 = 3 * (E_gp_b*Ia_m2) * cosd(160 - I_ap2_a); // // Internal developed power in W\n",
+"// 160 + I_ap2_a is the angle between E_gp_b  and I_ap2\n",
+"Pd2 = abs(P_d2); // Consider absolute value of power in W for calculating hp\n",
+"\n",
+"Horse_power2 = Pd2 / 746 ; // Horsepower developed by the armature in hp \n",
+"\n",
+"// case c\n",
+"E_rc = (V_p - E_gp_c * cosd(alpha)) + %i*(E_gp_c * sind(alpha));\n",
+"E_rc_m = abs(E_rc);//E_rc_m=magnitude of E_rc in volt\n",
+"E_rc_a = atan(imag(E_rc) /real(E_rc))*180/%pi;//E_rc_a=phase angle of E_rc in degrees\n",
+"\n",
+"I_ap3  = E_rc / Z_s ; // Armature current/phase in A/phase\n",
+"I_ap3_m = abs(I_ap3);//I_ap3_m=magnitude of I_ap3 in A\n",
+"I_ap3_a = atan(imag(I_ap3) /real(I_ap3))*180/%pi;//I_ap3_a=phase angle of I_ap3 in degrees \n",
+"\n",
+"cos_theta_c = cosd(I_ap3_a); // Power factor\n",
+"Ia_m3 = abs(I_ap3_m); // Absoulte value of magnitude of I_ap3\n",
+"\n",
+"P_d3 = 3 * (E_gp_c*Ia_m3) * cosd(160 - I_ap3_a); // // Internal developed power in W\n",
+"// 160 + I_ap3_a is the angle between E_gp_c  and I_ap3\n",
+"Pd3 = abs(P_d3); // Consider absolute value of power in W for calculating hp\n",
+"\n",
+"Horse_power3 = Pd3 / 746 ; // Horsepower developed by the armature in hp \n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-3 Solution : ');\n",
+"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 ')\n",
+"printf(' \n a: V_p = %.f <0 V \n ',V_p);\n",
+"printf(' \n    E_r in V = ');disp(E_ra);\n",
+"printf(' \n    E_r = %.2f <%.2f V \n ',E_ra_m,E_ra_a);\n",
+"printf(' \n    I_ap in A = ');disp(I_ap1);\n",
+"printf(' \n    I_ap = %.2f <%.2f A \n', I_ap1_m , I_ap1_a );\n",
+"printf(' \n    cos(theta) = %.4f lagging \n ', cos_theta_a );\n",
+"printf(' \n    P_d = %d W drawn from bus(motor operation)\n', P_d1 );\n",
+"printf(' \n    Horsepower = %.1f hp \n\n', Horse_power1 );\n",
+"\n",
+"printf(' \n b: E_r in V = ');disp(E_rb);\n",
+"printf(' \n    E_r = %.2f <%.2f V \n ',E_rb_m,E_rb_a);\n",
+"printf(' \n    I_ap in A = ');disp(I_ap2);\n",
+"printf(' \n    I_ap = %.2f <%.2f A \n', I_ap2_m , I_ap2_a );\n",
+"printf(' \n    cos(theta) = %.4f = %.f(unity PF) \n ', cos_theta_b, cos_theta_b );\n",
+"printf(' \n    P_d = %d W drawn from bus(motor operation)\n', P_d2 );\n",
+"printf(' \n    Horsepower = %.1f hp \n\n', Horse_power2 );\n",
+"\n",
+"printf(' \n c: E_r in V = ');disp(E_rc);\n",
+"printf(' \n    E_r = %.2f <%.2f V \n ',E_rc_m,E_rc_a);\n",
+"printf(' \n    I_ap in A = ');disp(I_ap3);\n",
+"printf(' \n    I_ap = %.2f <%.2f A \n', I_ap3_m , I_ap3_a );\n",
+"printf(' \n    cos(theta) = %.4f leading \n ', cos_theta_c );\n",
+"printf(' \n    P_d = %d W drawn from bus(motor operation)\n', P_d3 );\n",
+"printf(' \n    Horsepower = %.1f hp \n\n', Horse_power3 );\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.4: calculate_IL_Iap_Zp_IaZp_theta_deba_Egp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Y-connected synchronous dynamo\n",
+"P = 2 ; // No. of poles\n",
+"hp = 1000 ; // power rating of the synchronous motor in hp\n",
+"V_L = 6000 ; // Line voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"R_a = 0.52 ; // Effective armature resistance in ohm\n",
+"X_s = 4.2 ; // Synchronous reactance in ohm\n",
+"P_t = 811 ; // Input power in kW\n",
+"PF = 0.8 ; // Power factor leading\n",
+"\n",
+"// Calculations\n",
+"V_p = V_L / sqrt(3); // Phase voltage in volt\n",
+"\n",
+"// case a\n",
+"cos_theta = PF ; // Power factor leading\n",
+"I_L = (P_t*1000) / ( sqrt(3) * V_L * cos_theta); // Line armature current in A\n",
+"I_ap = I_L ; // Phase armature current in A\n",
+"\n",
+"// case b\n",
+"Z_p = R_a + %i * X_s ; // Impedance per phase in ohm\n",
+"Z_p_m = abs(Z_p);//Z_p_m=magnitude of Z_p in ohm\n",
+"Z_p_a = atan(imag(Z_p) /real(Z_p))*180/%pi;//Z_p_a=phase angle of Z_p in degrees\n",
+"\n",
+"// case c\n",
+"Ia_Zp = I_L * Z_p_m ; \n",
+"E_r = Ia_Zp ; \n",
+"\n",
+"// case d\n",
+"theta = acosd(0.8); // Power factor angle in degrees\n",
+"\n",
+"// case e\n",
+"funcprot(0); // Use  to avoid this message 'Warning : redefining function: beta'                    . \n",
+"beta = Z_p_a ; // \n",
+"deba = beta + theta // Difference angle at 0.8 leading PF in degrees\n",
+"\n",
+"// case f\n",
+"//  Generated voltage/phase in volt \n",
+"E_gp_f = sqrt( (E_r)^2 + (V_p)^2 - 2*E_r*V_p*cosd(deba) );\n",
+"\n",
+"// case g\n",
+"//  Generated voltage/phase in volt \n",
+"E_gp_g = ( V_p + Ia_Zp * cosd(180-deba) ) + %i * ( Ia_Zp * sind(180-deba) ); \n",
+"E_gp_g_m = abs(E_gp_g);//E_gp_g_m=magnitude of E_gp_g in volt\n",
+"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\n",
+"\n",
+"// case h\n",
+"IaZp = Ia_Zp * expm(%i * Z_p_a * (%pi/180) ); // voltage generated by alternator 1 in volt\n",
+"IaZp_m = abs(IaZp);//IaZp_m=magnitude of IaZp in A\n",
+"IaZp_a = atan(imag(IaZp) /real(IaZp))*180/%pi;//IaZp_a=phase angle of IaZp in degrees\n",
+"IaRa = IaZp_m*cosd(IaZp_a); // Real part of IaZp\n",
+"IaXs = IaZp_m*sind(IaZp_a); // Imaginery part of IaZp\n",
+"\n",
+"cos_theta = PF ; //\n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"//  Generated voltage/phase in volt \n",
+"E_gp_h = ( V_p * cos_theta - IaRa ) + %i * ( V_p * sin_theta + IaXs);\n",
+"E_gp_h_m = abs(E_gp_h);//E_gp_h_m=magnitude of E_gp_h in volt\n",
+"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\n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-4 Solution : ');\n",
+"printf(' \n a: I_L = %.2f \n    I_ap = %.2f A \n', I_L, I_ap );\n",
+"\n",
+"printf(' \n b: Z_p in ohm = ');disp(Z_p);\n",
+"printf(' \n    Z_p = %.3f <%.2f ohm \n ', Z_p_m , Z_p_a );\n",
+"\n",
+"printf(' \n c: IaZp = %.1f V \n    E_r = %.1f V \n ',Ia_Zp , E_r );\n",
+"\n",
+"printf(' \n d: Power factor angle,\n    theta = %.2f degrees leading \n ', theta );\n",
+"\n",
+"printf(' \n e: Difference angle,\n    deba = %.2f degrees \n ', deba );\n",
+"\n",
+"printf(' \n f: E_gp = %.f V \n ', E_gp_f );\n",
+" \n",
+"printf(' \n g: E_gp in V = ');disp(E_gp_g );\n",
+"printf(' \n    E_gp = %d <%.2f V \n',E_gp_g_m , E_gp_g_a );\n",
+"\n",
+"printf(' \n h: E_gp in V = ');disp(E_gp_h);\n",
+"printf(' \n    E_gp = %.f <%.2f V',E_gp_h_m, E_gp_h_a );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.5: calculate_torque_angle.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// Y-connected synchronous dynamo\n",
+"P = 2 ; // No. of poles\n",
+"hp = 1000 ; // power rating of the synchronous motor in hp\n",
+"V_L = 6000 ; // Line voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"R_a = 0.52 ; // Effective armature resistance in ohm\n",
+"X_s = 4.2 ; // Synchronous reactance in ohm\n",
+"P_t = 811 ; // Input power in kW\n",
+"PF = 0.8 ; // Power factor leading\n",
+"\n",
+"// Calculated values\n",
+"E_gp = 3687 ; // Generated voltage/phase in volt \n",
+"V_p = V_L / sqrt(3); // Phase voltage in volt\n",
+"E_r = 412.8 ; // Resultant EMF across armature/phase in volt\n",
+"deba = 119.81 ; // Difference angle at 0.8 leading PF in degrees\n",
+"theta = 36.87 ; // Power factor angle in degrees\n",
+"IaXs = 409.7 ; // Voltage drop across synchronous reactance in volt\n",
+"IaRa = 50.74 ; // Voltage drop across armature resistance in volt\n",
+"\n",
+"// Calculations\n",
+"\n",
+"// Torque angle alpha in degrees calculated by different Eqns\n",
+"// case a\n",
+"alpha1 = acosd( ( E_gp^2 + V_p^2 - E_r^2 ) / ( 2*E_gp*V_p ) ); // Eq.8-12\n",
+"\n",
+"// case b\n",
+"alpha2 = asind( ( E_r * sind(deba) ) / ( E_gp ) ); // Eq.8-13\n",
+"\n",
+"// case c\n",
+"alpha3 = theta - atand( (V_p*sind(theta) + IaXs) / (V_p*cosd(theta) - IaRa) );// Eq.8-14\n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-5 Solution : ');\n",
+"printf(' \n a: Using Eq.(8-12) \n    alpha = %.2f degrees \n ', alpha1 );\n",
+"\n",
+"printf(' \n b: Using Eq.(8-13) \n    alpha = %.2f degrees \n ', alpha2 );\n",
+"\n",
+"printf(' \n c: Using Eq.(8-14) \n    alpha = %.2f degrees \n ', alpha3 );\n",
+"printf(' \n    Slight variation in case c alpha is due to tan inverse value ');\n",
+"printf(' \n    which was calulated to be 42.445604 degrees, instead of 42.44 degrees(textbook).')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.6: calculate_Pp_Pt_hp_internal_and_external_torque_and_motor_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data as per Example 8-4\n",
+"// Y-connected synchronous dynamo\n",
+"P = 2 ; // No. of poles\n",
+"hp = 1000 ; // power rating of the synchronous motor in hp\n",
+"V_L = 6000 ; // Line voltage in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"R_a = 0.52 ; // Effective armature resistance in ohm\n",
+"X_s = 4.2 ; // Synchronous reactance in ohm\n",
+"P_t = 811 ; // Input power in kW\n",
+"PF = 0.8 ; // Power factor leading\n",
+"\n",
+"// Calculated values from Example 8-4\n",
+"E_gp = 3687 ; // Generated voltage/phase in volt \n",
+"\n",
+"I_a = 97.55 ; // Phase armature current in A\n",
+"\n",
+"phi = (42.45 - 0); // Phase angle between E_gp and I_a in degrees \n",
+"// where 42.45 and 0 are phase angles of E_gp and I_a in degrees respectively.\n",
+"\n",
+"// Calculations\n",
+"// case a \n",
+"P_p = E_gp * I_a * cosd(phi) / 1000; // Mechanical power developed per phase in kW\n",
+"\n",
+"P_t_a = 3 * P_p ; // Total mechanical power developed in kW\n",
+"\n",
+"// case b\n",
+"P_t_b = P_t_a / 0.746 ; // Internal power developed in hp at rated load\n",
+"\n",
+"// case c\n",
+"S = 120 * f / P ; // Speed of the motor in rpm\n",
+"T_int = ( P_t_b * 5252 ) / S ; // Internal torque developed in lb-ft\n",
+"\n",
+"// case d\n",
+"T_ext = ( hp * 5252 ) / 3600 ; // External torque developed in lb-ft\n",
+"eta = (T_ext / T_int) * 100 ; // Motor efficiency in percent\n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-6 Solution : ');\n",
+"printf(' \n a: Similar to a dc motor, the mechanical power developed in the armature');\n",
+"printf(' \n    is the product of the induced EMF per phase, the armature current');\n",
+"printf(' \n    per phase, and the cosine of the angle between them.\n');\n",
+"printf(' \n    P_p = %.3f kW \n    P_t = %.1f kW \n', P_p, P_t_a );\n",
+"\n",
+"printf(' \n b: P_t = %.1f hp \n ', P_t_b );\n",
+"\n",
+"printf(' \n c: T_int = %.f lb-ft \n ', T_int );\n",
+"\n",
+"printf(' \n d: T_ext = %d lb-ft  \n', T_ext );\n",
+"printf(' \n    Motor Efficiency,\n    eta = %.1f percent ', eta );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.7: EX8_7.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 2000 ; // Total power consumed by a factory in kW from the transformer\n",
+"cos_theta = 0.6 ; // 0.6 lagging power factor at which power is consumed -\n",
+"// - from the transformer\n",
+"sin_theta = sqrt(1 - (cos_theta)^2);\n",
+"theta = -acosd(0.6); // power factor angle at which power is consumed -\n",
+"// - from the transformer in degrees\n",
+"\n",
+"V_L = 6000 ; // Primary line voltage of a transformer in volt\n",
+"\n",
+"P = 750 ; // kW expected to be delivered by the dc motor-generator\n",
+"\n",
+"hp = 1000 ; // hp rating of the motor(induction or synchronous)\n",
+"V_L_m = 6000 ; // Line voltage of a synchronous(or induction) motor in volt\n",
+"cos_theta_sm = 0.8 ; // 0.8 leading power factor of the synchronous motor\n",
+"theta_sm = acosd(0.8); // power factor angle of the synchronous motor in degrees\n",
+"\n",
+"cos_theta_im = 0.8 ; // 0.8 lagging power factor of the induction motor\n",
+"theta_im = -acosd(0.8); // power factor angle of the induction motor in degrees\n",
+"\n",
+"eta = 0.92 ; // Efficiency of each motor\n",
+"\n",
+"// Calculations\n",
+"// case a : using Induction Motor(IM)\n",
+"P_m = ( hp * 746 ) / eta ; // Induction(or synchronous) motor load in W\n",
+"I_1 = P_m / ( sqrt(3) * V_L_m * cos_theta_im ); // Lagging current drawn by IM in A\n",
+"\n",
+"I_1_prime = P_o * 1000 / ( sqrt(3) * V_L * cos_theta ); // Original lagging  -\n",
+"// - factory load current in A\n",
+"\n",
+"// Total load current in A using Induction Motor :\n",
+"I_TM = I_1*(cosd(theta_im) + %i*sind(theta_im)) + I_1_prime*(cosd(theta) + %i*sind(theta)) ; \n",
+"I_TM_m = abs(I_TM);//I_TM_m = magnitude of I_TM in A\n",
+"I_TM_a = atan(imag(I_TM) /real(I_TM))*180/%pi;//I_TM_a=phase angle of I_TM in degrees\n",
+"\n",
+"PF_im = cosd(I_TM_a); // Overall PF using induction motor\n",
+"\n",
+"// case b: using synchronous motor\n",
+"I_s1 = P_m / ( sqrt(3) * V_L_m * cos_theta_sm ); // Lagging current drawn by IM in A\n",
+"\n",
+"// Total load current in A using synchronous motor :\n",
+"I_TSM = I_s1*(cosd(theta_sm) + %i*sind(theta_sm)) + I_1_prime*(cosd(theta) + %i*sind(theta)) ; \n",
+"I_TSM_m = abs(I_TSM);//I_TSM_m = magnitude of I_TSM in A\n",
+"I_TSM_a = atan(imag(I_TSM) /real(I_TSM))*180/%pi;//I_TSM_a=phase angle of I_TSM in degrees\n",
+"\n",
+"PF_sm = cosd(I_TSM_a); // Overall PF using Synchronous motor\n",
+"\n",
+"// case c \n",
+"percent_I_L = ( I_TM_m - I_TSM_m ) / I_TM_m * 100 ; // Percent reduction in  -\n",
+"// - total load current in percent\n",
+"\n",
+"// Display the results\n",
+"printf('Note : case a,I1 calculated is around 97.53 A instead of 47.53 A(textbook).\n')\n",
+"printf(' Note : case b,Actual I_s1 imaginary part is around 58.52 instead of ');\n",
+"printf(' \n        52.52(textbook)so slight variation in I_TSM and percent ')\n",
+"printf(' \n        reduction in total load current.\n')\n",
+"\n",
+"disp('Example 8-7 Solution : ');\n",
+"printf(' \n a: Induction(or sunchronous) motor load');\n",
+"printf(' \n    P_m = %.f W ',P_m);\n",
+"printf(' \n    Lagging current drawn by the IM = I1');\n",
+"printf(' \n    I_1 = %.2f <-%.2f A \n',I_1,acosd(cos_theta_sm));\n",
+"printf(' \n    I_1 in A = ');disp(I_1*cosd(-36.87)+%i*I_1*sind(-36.87));\n",
+"printf(' \n    Original lagging factory load current = I_1_prime');\n",
+"printf(' \n    I_1_prime in A = ');disp(I_1_prime*cosd(theta)+%i*I_1_prime*sind(theta));\n",
+"printf(' \n    I_1_prime = %.1f <-%.2f A \n',I_1_prime,acosd(cos_theta));\n",
+"printf(' \n    Total load current = motor load + factory load');\n",
+"printf(' \n    I_TM = I_1 + I_1_prime\n');\n",
+"printf(' \n    I_TM in A = ');disp(I_TM);\n",
+"printf(' \n    I_TM = %.1f <%.1f A \n ',I_TM_m , I_TM_a );\n",
+"printf(' \n    Overall system PF = %.4f lagging \n ', PF_im );\n",
+"\n",
+"printf(' \n b: Synchronous motor load\n    I_s1 = %.2f <%.2f A\n',I_1,acosd(cos_theta_sm));\n",
+"printf(' \n    I_s1 in A = ');disp(I_s1*cosd(36.87)+%i*I_s1*sind(36.87));\n",
+"printf(' \n    Total load current : I_TSM = I_s1 + I_1_prime \n');\n",
+"printf(' \n    I_TSM in A = ');disp(I_TSM);\n",
+"printf(' \n    I_TSM = %.1f <%.1f A \n ',I_TSM_m , I_TSM_a );\n",
+"printf(' \n    Overall system PF = %.1f lagging \n ', PF_sm );\n",
+"\n",
+"printf(' \n c: Percent reduction in total load current = %.1f percent \n',percent_I_L);\n",
+"\n",
+"printf(' \n d: PF improvement: Using the synchronous motor ( in lieu of the IM)');\n",
+"printf(' \n    raises the total system PF from %.4f lagging to %.1f lagging.',PF_im,PF_sm);\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.8: calculate_Tp_and_hp.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data from Ex.8-3a\n",
+"// 3- phase Y-connected synchronous motor\n",
+"P = 6 ; // No. of poles\n",
+"hp = 50 ; // power rating of the synchronous motor in hp\n",
+"V_L = 440 ; // Line voltage in volt\n",
+"X_s = 2.4 ; // Synchronous reactance in ohm\n",
+"R_a = 0.1 ; // Effective armature resistance in ohm\n",
+"alpha = 20 ; // The rotor shift from the synchronous position in\n",
+"// electrical degrees.\n",
+"E_gp = 240 ; // Generated voltage/phase in volt when the motor is under-excited\n",
+"f = 60 ; // Frequency in Hz\n",
+"\n",
+"// Calculated values from Example 8-3a\n",
+"V_p = 254 ; // Phase voltage in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"// Torque developed per phase Using Eq.(8-17a)\n",
+"S = 120 * f / P ; // Speed of the motor in rpm\n",
+"T_p = ( 7.04 * E_gp * V_p ) / ( S*X_s) * sind(alpha); \n",
+"\n",
+"// case b\n",
+"// Total horsepower developed using part a\n",
+"Horsepower = ( 3*T_p*S )/5252; \n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-8 Solution : ');\n",
+"printf(' \n From given and calculated data of Ex.8-3a,\n');\n",
+"printf(' \n a: T_p = %.2f lb-ft \n ', T_p );\n",
+"\n",
+"printf(' \n b: Horsepower = %.1f hp ', Horsepower );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 8.9: EX8_9.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS\n",
+"// Example 8-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P_o = 2000 ; // Total power consumed by a factory in kW \n",
+"cos_theta = 0.6 ; // 0.6  power factor at which power is consumed \n",
+"sin_theta = sqrt( 1 - (cos_theta)^2 );\n",
+"V = 6000 ; // Line voltage in volt\n",
+"// Synchronous capacitor is used to raise the overall PF to unity\n",
+"P_loss_cap = 275 ; // Synchronous capacitor losses in kW\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"S_o_conjugate = P_o / cos_theta ; // apparent complex power in kW\n",
+"jQ_o = S_o_conjugate * sin_theta ; // Original kilovars of lagging load\n",
+"\n",
+"// case b\n",
+"jQ_c = -jQ_o ; // Kilovars of correction needed to bring the PF to unity\n",
+"\n",
+"// case c\n",
+"R = P_loss_cap ; // Synchronous capacitor losses in kW\n",
+"S_c_conjugate = R - %i*( abs(jQ_c) ) ; // kVA rating of the synchronous capacitor\n",
+"S_c_conjugate_m = abs(S_c_conjugate);//S_c_conjugate_m = magnitude of S_c_conjugate in kVA\n",
+"S_c_conjugate_a = atan(imag(S_c_conjugate) /real(S_c_conjugate))*180/%pi;\n",
+"//S_c_conjugate_a=phase angle of S_c_conjugate in degrees \n",
+"PF = cosd(S_c_conjugate_a); // Power factor of the synchronous capacitor\n",
+"\n",
+"// case d\n",
+"I_o = S_o_conjugate * 1000 / V ; // Original current drawn from the mains in A\n",
+"\n",
+"\n",
+"// case e\n",
+"P_f = P_o + P_loss_cap ; // Total power in kW\n",
+"S_f = P_f ; // Total apparent power in kW\n",
+"S_f_m = abs(S_f);//S_f_m = magnitude of S_f in A\n",
+"S_f_a = atan(imag(S_f) /real(S_f))*180/%pi;//S_f_a=phase angle of S_f in degrees \n",
+"\n",
+"I_f = S_f * 1000 / V ; // Final current drawn from the mains after correction in A\n",
+"\n",
+"// Display the results\n",
+"disp('Example 8-9 Solution : ');\n",
+"printf(' \n a: S*o = %d kVA \n', S_o_conjugate );\n",
+"printf(' \n    +jQo in kvar = ');disp(%i*jQ_o);\n",
+"\n",
+"printf(' \n b: -jQc in kvar =  ' );disp(%i*jQ_c);\n",
+"\n",
+"printf(' \n c: S*c in kVA = ');disp(S_c_conjugate);\n",
+"printf(' \n    S*c = %.f <%.1f kVA \n', S_c_conjugate_m , S_c_conjugate_a );\n",
+"printf(' \n    PF = %.3f leading \n',PF );\n",
+"\n",
+"printf(' \n d: I_o = %.1f A \n ',I_o );\n",
+"\n",
+"printf(' \n e: S_f in A = ');disp(S_f);\n",
+"printf(' \n    S_f = %d <%d kVA \n' , S_f_m , S_f_a );\n",
+"printf(' \n    I_f = %.1f A \n ', I_f);\n",
+"\n",
+"printf(' \n f: See Fig.8-25.');"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}
diff --git a/Electric_Machinery_And_Transformers_by_I_L_Kosow/9-POLYPHASE_INDUCTION_OR_ASYNCHRONOUS_DYNAMOS.ipynb b/Electric_Machinery_And_Transformers_by_I_L_Kosow/9-POLYPHASE_INDUCTION_OR_ASYNCHRONOUS_DYNAMOS.ipynb
new file mode 100644
index 0000000..d9a687c
--- /dev/null
+++ b/Electric_Machinery_And_Transformers_by_I_L_Kosow/9-POLYPHASE_INDUCTION_OR_ASYNCHRONOUS_DYNAMOS.ipynb
@@ -0,0 +1,1360 @@
+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 9: POLYPHASE INDUCTION OR ASYNCHRONOUS DYNAMOS"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.10: calculate_Elr_Ir_Pin_RCL_RPD_torques.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-10\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 4 ; // Number of poles in WRIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"V = 220 ; // Line voltage in volt\n",
+"V_p = 220 ; // Phase voltage in volt (delta connection)\n",
+"hp_WRIM = 1 ; // Power rating of WRIM in hp\n",
+"S_r = 1740 ; // Full-load rated speed in rpm \n",
+"R_r = 0.3 ; // rotor resistance per phase in ohm/phase \n",
+"R_x = 0.7 ; // Added resistance in ohm/phase \n",
+"X_lr =1 ; // Locked rotor reactance in ohm\n",
+"\n",
+"// Calculations\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// case a\n",
+"E_lr = V_p / 4 ; // Locked-rotor voltage per phase\n",
+"\n",
+"// case b\n",
+"s = ( S - S_r)/S ; // slip\n",
+"I_r = E_lr / sqrt( (R_r/s)^2 + (X_lr)^2 ); // Rotor current per phase at rated speed\n",
+"\n",
+"// case c\n",
+"P_in = ((I_r)^2 * R_r)/s ; // Rated rotor power input per phase\n",
+"\n",
+"// case d\n",
+"P_RL = (I_r)^2 * R_r ; // Rated copper loss per phase\n",
+"\n",
+"// case e\n",
+"P_d_W = P_in - P_RL ; // Rotor power developed per phase in W\n",
+"P_d_hp = P_d_W/746 ; // Rotor power developed per phase in hp\n",
+"\n",
+"// case f\n",
+"hp = P_d_hp ; // Rotor power developed per phase in hp\n",
+"T_d1 = (hp*5252)/S_r ; // Rotor torque developed in lb-ft per phase by method 1\n",
+"T_d2 = 7.04*(P_in/S) ; // Rotor torque developed in lb-ft per phase by method 2\n",
+"\n",
+"T_dm = 3*T_d1 ; // Total rotor torque in lb-ft\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-10 Solution : ');\n",
+"printf(' \n a: Locked-rotor voltage per phase : \n    E_lr = %d V \n ',E_lr);\n",
+"\n",
+"printf(' \n b: slip : \n    s = %.2f \n',s);\n",
+"printf(' \n    Rotor current per phase at rated speed:\n    I_r = %.3f A/phase \n ',I_r);\n",
+"\n",
+"printf(' \n c: Rated rotor power input per phase :\n    P_in = %d W/phase \n ',P_in);\n",
+"\n",
+"printf(' \n d: Rated copper loss per phase : \n    P_RL = %.2f W \n ',P_RL);\n",
+"\n",
+"printf(' \n e: Rotor power developed per phase in W :\n    P_d = %.1f W/phase ',P_d_W);\n",
+"printf(' \n\n    Rotor power developed per phase in hp :\n    P_d = %.2f hp/phase \n ',P_d_hp);\n",
+"\n",
+"printf(' \n f: Rotor torque developed in lb-ft per phase :\n    T_d = %.1f lb-ft  (method 1)',T_d1);\n",
+"printf(' \n\n    T_d = %.1f lb-ft  (method 2)',T_d2);\n",
+"printf(' \n\n    Total rotor torque : \n    T_dm = %.1f lb-ft )\n ',T_dm);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.11: calculate_Elr_Ir_Pin_RCL_RPD_and_torques.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-11\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"// 3-phase WRIM\n",
+"V_L = 208 ; // Voltage rating of the WRIM in volt\n",
+"P = 6 ; // Number of poles in WRIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"P_o = 7.5 ; // Power rating of WRIM in hp\n",
+"S_r = 1125 ; // Full-load rotor speed in rpm\n",
+"R_r = 0.08 ; // Rotor resistance in ohm/phase \n",
+"X_lr = 0.4 ; // Locked rotor resistance in ohm/phase\n",
+"\n",
+"// Calculations\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// case a\n",
+"E_lr = (V_L / sqrt(3))/2 ; // Locked rotor voltage per phase\n",
+"\n",
+"// case b\n",
+"s = (S - S_r)/S ; // Full-load rated slip\n",
+"I_r = E_lr / sqrt( (R_r/s)^2 + (X_lr)^2 ); // Rotor current in A per phase at rated speed\n",
+"\n",
+"// case c\n",
+"P_in = ( (I_r)^2 * R_r )/s ; // Rated rotor power input per phase in (W/phase)\n",
+"\n",
+"// case d\n",
+"P_RL = ( (I_r)^2 * R_r ); // Rated rotor copper loss per phase (in W/phase)\n",
+"\n",
+"// case e\n",
+"// Subscript W in P_d indicates calculating P_d in W\n",
+"P_d_W = P_in - P_RL ; // Rotor power developed per phase (in W/phase)\n",
+"// Subscript hp in P_d indicates calculating P_d in hp\n",
+"P_d_hp = P_d_W/746 ; // Rotor power developed per phase (in hp/phase)\n",
+"\n",
+"// case f\n",
+"// subscript 1 in T_d indicates method 1 for calculating T_d\n",
+"hp = P_d_hp ;\n",
+"T_d1 = (hp*5252)/S_r ; // Rotor torque developed per phase in lb-ft\n",
+"\n",
+"// subscript 2 in T_d indicates method 2 for calculating T_d\n",
+"T_d2 = 7.04*(P_in/S); // Rotor torque developed per phase in lb-ft\n",
+"\n",
+"// case g\n",
+"T_dm = 3*T_d1 ; // Total rotor torque in lb-ft\n",
+"\n",
+"// case h\n",
+"T_o = 7.04*(P_o*746)/S_r ; // Total output rotor torque in lb-ft\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-11 Solution : ');\n",
+"\n",
+"printf(' \n    Note: Slight variations in the answers I_r,P_in,P_RL,P_d,T_d ');\n",
+"printf(' \n          are because of non-approximation of E_lr and (R_r/s)^2 + (X_lr)^2');\n",
+"printf(' \n          while calulating in scilab.\n');\n",
+"\n",
+"printf(' \n a: Locked rotor voltage per phase :\n    E_lr = %d V\n',E_lr);\n",
+"\n",
+"printf(' \n b: slip :\n    s = %.4f ',s);\n",
+"printf(' \n\n    Rotor current per phase at rated speed :\n    I_r = %.2f A/phase\n',I_r);\n",
+"\n",
+"printf(' \n c: Rated rotor power input per phase :\n    P_in = %.f W/phase\n',P_in);\n",
+"\n",
+"printf(' \n d: Rated rotor copper loss per phase :\n    P_RL = %.1f W/phase\n',P_RL);\n",
+"\n",
+"printf(' \n e: Rotor power developed per phase ');\n",
+"printf(' \n    P_d = %.f W/phase \n    P_d = %.2f hp/phase\n',P_d_W,P_d_hp);\n",
+"\n",
+"printf(' \n f: Rotor torque developed per phase : ');\n",
+"printf(' \n    (method 1)\n    T_d = %.1f lb-ft/phase',T_d1);\n",
+"printf(' \n\n    (method 2)\n    T_d = %.1f lb-ft/phase\n',T_d2);\n",
+"\n",
+"printf(' \n g: Total rotor torque : \n    T_dm = %d lb-ft\n',T_dm);\n",
+"\n",
+"printf(' \n h: Total output rotor torque : \n    T_o = %d lb-ft',T_o);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.12: calculate_s_and_Sr_for_Tmax.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-12\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data as per Ex.9-10\n",
+"P = 4 ; // Number of poles in WRIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"V = 220 ; // Line voltage in volt\n",
+"V_p = 220 ; // Phase voltage in volt (delta connection)\n",
+"hp_WRIM = 1 ; // Power rating of WRIM in hp\n",
+"S_r = 1740 ; // Full-load rated speed in rpm \n",
+"R_r = 0.3 ; // rotor resistance per phase in ohm/phase \n",
+"R_x = 0.7 ; // Added resistance in ohm/phase \n",
+"X_lr = 1 ; // Locked rotor reactance in ohm\n",
+"// Calculations from Ex.9-10\n",
+"E_lr = V_p / 4 ; // Locked-rotor voltage per phase\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// Calculations (Ex.9-12)\n",
+"P_in = (E_lr)^2 / (2*X_lr); // rotor power input(RPI) in W/phase\n",
+"P_in_total = P_in * 3 ; // Total 3-phase rotor power input(RPI) in W\n",
+"T_max = 7.04*(P_in_total/S); // Maximum torque developed in lb-ft\n",
+"s_b = R_r / X_lr ; // Slip\n",
+"s = s_b;\n",
+"S_r = S*(1 - s); // Rotor speed in rpm for T_max\n",
+"// Display the results\n",
+"disp('Example 9-12 Solution : ');\n",
+"printf(' \n Rotor power input (RPI) per phase is : ');\n",
+"printf(' \n P_in = %.1f W/phase \n',P_in);\n",
+"printf(' \n The total 3-phase rotor power input (RPI) is : ');\n",
+"printf(' \n P_in = %.1f W\n',P_in_total);\n",
+"printf(' \n Substituting in Eq.(9-19),\n T_max = %.2f lb-ft\n',T_max);\n",
+"printf(' \n Then, s_b = %.1f \n and S_r = %d rpm',s_b,S_r);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.13: calculate_starting_torque.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-13\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data as per Ex.9-10\n",
+"P = 4 ; // Number of poles in WRIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"V = 220 ; // Line voltage in volt\n",
+"V_p = 220 ; // Phase voltage in volt (delta connection)\n",
+"hp_WRIM = 1 ; // Power rating of WRIM in hp\n",
+"S_r = 1740 ; // Full-load rated speed in rpm \n",
+"R_r = 0.3 ; // rotor resistance per phase in ohm/phase \n",
+"R_x = 0.7 ; // Added resistance in ohm/phase \n",
+"X_lr = 1 ; // Locked rotor reactance in ohm\n",
+"// Calculations\n",
+"E_lr = V_p / 4 ; // Locked-rotor voltage per phase\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// Total 3-phase rotor power input(RPI) in W\n",
+"P_in = 3 * ( (E_lr)^2 ) / ( (R_r)^2 + (X_lr)^2 ) * R_r ;\n",
+"T_s = 7.04 * (P_in/S); // Starting torque developed in lb-ft\n",
+"// Display the results\n",
+"disp('Example 9-13 Solution : ');\n",
+"printf(' \n P_in = %.f W \n',P_in);\n",
+"printf(' \n From Eq.(9-19),starting torque is : \n T_s = %.2f lb-ft',T_s);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.14: calculate_full_load_and_starting_torques.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-14\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"T_max = 17.75 ; // Maximum torque developed in lb-ft\n",
+"s_max = 0.3 ; // Slip for which T_max occurs\n",
+"s_a = 0.0333 ; // slip (case a)\n",
+"s_b = 1.0 ; // slip (case b) \n",
+"// Calculations\n",
+"// Subscript a in T indicates case a\n",
+"T_a = T_max * ( 2 / ((s_max/s_a) + (s_a/s_max)) ); // Full-load torque in lb-ft\n",
+"// Subscript b in T indicates case b\n",
+"T_b = T_max * ( 2 / ((s_max/s_b) + (s_b/s_max)) ); // Starting torque in lb-ft\n",
+"// Display the results\n",
+"disp('Example 9-14 Solution : ');\n",
+"printf(' \n a: Full-load torque at slip = %.4f \n    T = %.1f lb-ft\n',s_a,T_a);\n",
+"printf(' \n b: Starting torque at slip = %.1f \n    T = %.2f lb-ft\n',s_b,T_b);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.15: EX9_15.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-15\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"// 3-phase Y-connected SCIM\n",
+"P = 4 ; // Number of poles in SCIM \n",
+"S_r = 1746 ; // Rated rotor speed in rpm\n",
+"V = 220 ; // Voltage rating of SCIM in volt\n",
+"f = 60 ; // Frequency in Hz\n",
+"P_hp = 10 ; // power rating of SCIM in hp\n",
+"R_a = 0.4 ; // Armature resistance in ohm\n",
+"R_r = 0.14 ; // Rotor resistance in ohm\n",
+"jXm = 16 ; // Reactance in ohm\n",
+"jXs = 0.35 ; // Synchronous reactance in ohm\n",
+"jXlr = 0.35 ; // Locked rotor reactance in ohm\n",
+"P_r_total = 360 ; // Total rotational losses in W\n",
+"// Calculations\n",
+"V_p = V / sqrt(3); // Voltage per phase in volt\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// preliminary calculations\n",
+"s = ( S - S_r)/S ; // slip\n",
+"disp('Example 9-15 :');\n",
+"printf(' \n From Fig.9-13,using the format method of mesh analysis,we may');\n",
+"printf(' \n write the array by inspection :\n');\n",
+"printf(' \n __________________________________________________________');\n",
+"printf(' \n    I_1(A) \t\t I_2(A) \t\t V(volt)');\n",
+"printf(' \n __________________________________________________________');\n",
+"printf(' \n   (0.4 + j16.35) \t -(0 + j16) \t\t (127 + j0)');\n",
+"printf(' \n   -(0 + j16) \t\t (4.67 + j16.35) \t     0');\n",
+"printf(' \n __________________________________________________________');\n",
+"A = [ (0.4 + %i*16.35) -%i*16 ; (-%i*16)  (4.67 + %i*16.35) ]; // Matrix containing above mesh eqns array\n",
+"delta = det(A); // Determinant of A\n",
+"// case a : Stator armature current I_p in A\n",
+"I_p = det( [ (127+%i*0) (-%i*16) ; 0 (4.67 + %i*16.35) ] ) / delta ;\n",
+"I_p_m = abs(I_p);//I_p_m=magnitude of I_p in A\n",
+"I_p_a = atan(imag(I_p) /real(I_p))*180/%pi;//I_p_a=phase angle of I_p in degrees\n",
+"I_1 = I_p ; // Stator armature current in A\n",
+"// case b : Rotor current I_r per phase in A\n",
+"I_r = det( [ (0.4 + %i*16.35) (127+%i*0) ; (-%i*16) 0 ] ) / delta ;\n",
+"I_r_m = abs(I_r);//I_r_m=magnitude of I_r in A\n",
+"I_r_a = atan(imag(I_r) /real(I_r))*180/%pi;//I_r_a=phase angle of I_r in degrees\n",
+"// case c\n",
+"theta_1 = I_p_a ; // Motor PF angle in degrees\n",
+"cos_theta1 = cosd(theta_1); // Motor PF\n",
+"// case d\n",
+"I_p = I_p_m ; // Stator armature current in A\n",
+"SPI = V_p * I_p * cos_theta1 ; // Stator Power Input in W\n",
+"// case e\n",
+"SCL = (I_p)^2 * R_a ; // Stator Copper Loss in W\n",
+"// case f\n",
+"// Subscripts 1 and 2 for RPI indicates two methods of calculating RPI\n",
+"RPI_1 = SPI - SCL ; // Rotor Power Input in W\n",
+"RPI_2 = (I_r_m)^2 * (R_r/s); // Rotor Power Input in W\n",
+"RPI =RPI_1 ;\n",
+"// case g\n",
+"// Subscripts 1 , 2 and 3 for RPD indicates three methods of calculating RPD\n",
+"RPD_1 = RPI * ( 1 - s ); // Rotor Power Developed in W\n",
+"RCL = s*(RPI); // Rotor copper losses in W\n",
+"RPD_2 = RPI - RCL ; // Rotor Power Developed in W\n",
+"RPD_3 = (I_r_m)^2 * R_r * ((1-s)/s); // Rotor Power Developed in W\n",
+"RPD = RPD_1 ;\n",
+"// case h\n",
+"P_r = P_r_total / 3 ; // Rotational Losses per phase in W\n",
+"P_o = RPD - P_r ; // Rotor power per phase in W\n",
+"P_to = 3*P_o ; // Total rotor power in W\n",
+"// case i\n",
+"T = 7.04 * (P_to/S_r); // Total 3-phase torque in lb-ft\n",
+"// case j\n",
+"P_t = P_to ;\n",
+"hp = P_t / 746 ; // Output horsepower\n",
+"// case k\n",
+"P_in = SPI ; // Input power to stator in W\n",
+"eta = P_o / P_in * 100 ; // Motor efficiency at rated load\n",
+"// Display the results\n",
+"disp('Solution : ');\n",
+"printf(' \n Preliminary calculations\n');\n",
+"printf(' \n Slip : s = %.2f \n R_r/s = %.2f ohm \n',s,R_r/s);\n",
+"printf(' \n Determinant Δ = ');disp(delta);\n",
+"printf(' \n a: Stator armature current :\n    I_p in A = ');disp(I_1);\n",
+"printf(' \n    I_p = I_1 = %.2f <%.2f A \n ',I_p_m , I_p_a );\n",
+"printf(' \n b: Rotor current per phase :\n    I_r in A = ');disp(I_r);\n",
+"printf(' \n    I_r = I_2 = %.3f <%.2f A \n ',I_r_m , I_r_a );\n",
+"printf(' \n c: Motor PF :\n    cosӨ1 = %.4f \n',cos_theta1);\n",
+"printf(' \n d: Stator Power Input :\n    SPI = %d W \n',SPI);\n",
+"printf(' \n e: Stator Copper Loss :\n    SCL = %.f W \n',SCL);\n",
+"printf(' \n f: Rotor Power Input :\n    RPI = %d W(method 1) ', RPI_1);\n",
+"printf(' \n    RPI = %.f W (method 2)\n',RPI_2);\n",
+"printf(' \n    Note: RPI calculated by 2nd method slightly varies from that of');\n",
+"printf(' \n          textbook value because of non-approximation of I_r while');\n",
+"printf(' \n          calculating in scilab.\n')\n",
+"printf(' \n g: Rotor Power Developed :\n    RPD = %.f W \n',RPD_1);\n",
+"printf(' \n    Rotor copper loss :\n    RCL = %d W\n',RCL);\n",
+"printf(' \n    RPD = %.f W \n    RPD = %d W \n ',RPD_2,RPD_3);\n",
+"printf(' \n h: Rotor power per phase :\n    P_o/φ = %f W/φ ',P_o);\n",
+"printf(' \n\n    Total rotor power:\n    P_to = %f W \n',P_to);\n",
+"printf(' \n    Above P_o/φ and P_to values are not approximated while calculating in ');\n",
+"printf(' \n    SCILAB.So,they vary slightly from textbook values.\n');\n",
+"printf(' \n i: Total 3-phase output torque :\n    T = %.f lb-ft\n',T);\n",
+"printf(' \n j: Output horsepower : \n    hp = %.1f hp \n',hp);\n",
+"printf(' \n k: Motor efficiency at rated load :\n    η = %.1f percent \n',eta)\n",
+"printf(' \n Power flow diagram (per phase)\n');\n",
+"printf(' \n   SPI----------> RPI---------> RPD----------> P_o');\n",
+"printf(' \n (%d W)   |  (%d W)  |   (%d W)   |   (%d W)',SPI,RPI_1,RPD_3,P_o);\n",
+"printf(' \n            |            |              |');\n",
+"printf(' \n           SCL          RCL            P_r');\n",
+"printf(' \n         (%.f W)       (%d W)         (%d W)',SCL,RCL,P_r);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.16: calculate_Ism_IL_Ts_and_percent_IL_and_percent_Ts.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-16\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"// three-phase SCIM\n",
+"V = 208 ; // Rated voltage in volt\n",
+"P_o = 15 ; // Rated power in hp\n",
+"I = 42 ; // Rated current in A\n",
+"I_st = 252 ; // Starting current in A\n",
+"T_st = 120 ; // Full-voltage starting torque in lb-ft\n",
+"tap = 60*(1/100) ; // Tapping in % employed by compensator\n",
+"// Calculations\n",
+"// case a\n",
+"I_sm = tap * I_st ; // Motor starting current in A at reduced voltage \n",
+"// case b\n",
+"I_L = tap * I_sm ; // Motor line current in A(neglecting tarnsformer exciting \n",
+"// current and losses)\n",
+"// case c\n",
+"T_s = (tap)^2 * T_st ; // Motor starting torque at reduced voltage in lb-ft\n",
+"// case d\n",
+"percent_I_L = I_L / I_st * 100 ; // Percent line current at starting\n",
+"// case e\n",
+"percent_T_st = T_s / T_st * 100 ; // Percent motor starting torque\n",
+"// Display the results\n",
+"disp('Example 9-16 Solution : ');\n",
+"printf(' \n a: Motor starting current at reduced voltage : ');\n",
+"printf(' \n    I_sm = %.1f A to the motor.\n',I_sm);\n",
+"printf(' \n b: Motor line current neglecting tarnsformer exciting current and losses :');\n",
+"printf(' \n    I_L = %.2f A drawn from the mains.\n',I_L);\n",
+"printf(' \n c: Motor starting torque at reduced voltage :\n    T_s = %.1f lb-ft\n',T_s);\n",
+"printf(' \n d: Percent line current at starting : ');\n",
+"printf(' \n    = %.f percent of line current at full voltage.\n',percent_I_L);\n",
+"printf(' \n e: Percent motor starting torque : ');\n",
+"printf(' \n    = %d percent of starting torque at full voltage.\n',percent_T_st);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.17: calculate_T_s_Sr_for_different_V.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-17\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"// three-phase SCIM\n",
+"V_o = 220 ; // Rated voltage in volt\n",
+"P = 4 ; // Number of poles in SCIM\n",
+"P_o = 10 ; // Rated power in hp\n",
+"f = 60 ; // Frequency in Hz(assume,not given)\n",
+"T_o = 30 ; // Rated torque in lb-ft\n",
+"S_r = 1710 ; // Rated rotor speed in rpm\n",
+"V_n1 = 242 ; // Impressed stator voltage in volt(case a)\n",
+"V_n2 = 198 ; // Impressed stator voltage in volt(case b)\n",
+"// Calculations\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// case a : Impressed stator voltage = 242 V\n",
+"s_o = (S - S_r)/S ; // Rated slip\n",
+"T_n1 = T_o * (V_n1/V_o)^2 ; // New torque in lb-ft\n",
+"s_n1 = s_o * (T_o/T_n1); // New slip\n",
+"S_rn1 = S*(1 - s_n1); \n",
+"// case b : Impressed stator voltage = 198 V\n",
+"T_n2 = T_o * (V_n2/V_o)^2 ; // New torque in lb-ft\n",
+"s_n2 = s_o * (T_o/T_n2); // New slip\n",
+"S_rn2 = S*(1 - s_n2);\n",
+"// case c\n",
+"// Subscript a in percent_slip and percent_speed indicates part a\n",
+"percent_slip_a = (s_o - s_n1)/s_o * 100 ; // Percent change in slip in part(a)\n",
+"percent_speed_a = (S_rn1 - S_r)/S_r * 100; // Percent change in speed in part(a)\n",
+"// case d\n",
+"// Subscript b in percent_slip and percent_speed indicates part b\n",
+"percent_slip_b = (s_n2 - s_o)/s_o * 100 ; // Percent change in slip in part(b)\n",
+"percent_speed_b = (S_r - S_rn2)/S_r * 100; // Percent change in speed in part(b)\n",
+"// Display the results\n",
+"disp('Example 9-17 Solution : ');\n",
+"printf(' \n a: Rated slip :\n    s = %.2f\n',s_o);\n",
+"printf(' \n    For impressed stator voltage = %d V \n ',V_n1);\n",
+"printf(' \n    New torque :\n    T_n = %.1f lb-ft \n ',T_n1);\n",
+"printf(' \n    New slip :\n    s_n = %f \n ',s_n1);\n",
+"printf(' \n    New rotor speed :\n    S_r = %f rpm \n',S_rn1);\n",
+"printf(' \n b: For impressed stator voltage = %d V \n ',V_n2);\n",
+"printf(' \n    New torque :\n    T_n = %.1f lb-ft \n ',T_n2);\n",
+"printf(' \n    New slip :\n    s_n = %f \n ',s_n2);\n",
+"printf(' \n    New rotor speed :\n    S_r = %f rpm \n',S_rn2);\n",
+"printf(' \n c: Percent change in slip in part(a)');\n",
+"printf(' \n    = %.1f percent decrease.\n',percent_slip_a);\n",
+"printf(' \n    Percent change in speed in part(a)');\n",
+"printf(' \n    = %.2f percent increase \n',percent_speed_a);\n",
+"printf(' \n d: Percent change in slip in part(b)');\n",
+"printf(' \n    = %.2f percent increase.\n',percent_slip_b);\n",
+"printf(' \n    Percent change in speed in part(b)');\n",
+"printf(' \n    = %.2f percent decrease\n',percent_speed_b);\n",
+"printf(' \n   SLIGHT VARIATIONS IN PERCENT CHANGE IN SLIP AND SPEED ARE DUE TO');\n",
+"printf(' \n   NON-APPROXIMATION OF NEW SLIPS AND NEW SPEEDS CALCULATED IN SCILAB.')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.18: calculate_T_s_Sr_for_different_impressed_stator_V.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-18\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"// three-phase WRIM\n",
+"V_o = 220 ; // Rated voltage in volt\n",
+"P_o = 10 ; // Rated power in hp\n",
+"P = 4 ; // Number of poles in WRIM(assumption)\n",
+"f = 60 ; // Frequency in Hz(assume,not given)\n",
+"R_ro = 0.3 ; // Rotor resistance in ohm\n",
+"T_o = 30 ; // Rated torque in lb-ft\n",
+"S_r = 1750 ; // Rated rotor speed in rpm\n",
+"R_r_ext = 1.7 ; // External rotor resistance in ohm/phase inserted in the rotor ckt \n",
+"R_rn = R_ro + R_r_ext ; // Total rotor resistance in ohm\n",
+"V_n1 = 240 ; // Impressed stator voltage in volt(case a)\n",
+"V_n2 = 208 ; // Impressed stator voltage in volt(case b)\n",
+"V_n3 = 110 ; // Impressed stator voltage in volt(case c)\n",
+"// Calculations\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// case a : Impressed stator voltage = 240 V\n",
+"s_o = (S - S_r)/S ; // Rated slip\n",
+"T_n1 = T_o * (V_n1/V_o)^2 ; // New torque in lb-ft\n",
+"s_n1 = s_o * (T_o/T_n1) * (R_rn/R_ro); // New slip\n",
+"S_rn1 = S*(1 - s_n1); \n",
+"// case b : Impressed stator voltage = 208 V\n",
+"T_n2 = T_o * (V_n2/V_o)^2 ; // New torque in lb-ft\n",
+"s_n2 = s_o * (T_o/T_n2) * (R_rn/R_ro); // New slip\n",
+"S_rn2 = S*(1 - s_n2);\n",
+"// case c : Impressed stator voltage = 110 V\n",
+"T_n3 = T_o * (V_n3/V_o)^2 ; // New torque in lb-ft\n",
+"s_n3 = s_o * (T_o/T_n3) * (R_rn/R_ro); // New slip\n",
+"S_rn3 = S*(1 - s_n3);\n",
+"// Display the results\n",
+"disp('Example 9-18 Solution : ');\n",
+"printf(' \n a: Rated slip :\n    s = %f\n',s_o);\n",
+"printf(' \n    For impressed stator voltage = %d V \n ',V_n1);\n",
+"printf(' \n    New torque :\n    T_n = %.1f lb-ft \n ',T_n1);\n",
+"printf(' \n    New slip :\n    s_n = %f \n ',s_n1);\n",
+"printf(' \n    New rotor speed :\n    S_r = %f rpm \n',S_rn1);\n",
+"printf(' \n b: For impressed stator voltage = %d V \n ',V_n2);\n",
+"printf(' \n    New torque :\n    T_n = %.2f lb-ft \n ',T_n2);\n",
+"printf(' \n    New slip :\n    s_n = %f \n ',s_n2);\n",
+"printf(' \n    New rotor speed :\n    S_r = %f rpm \n',S_rn2);\n",
+"printf(' \n c: For impressed stator voltage = %d V \n ',V_n3);\n",
+"printf(' \n    New torque :\n    T_n = %.1f lb-ft \n ',T_n3);\n",
+"printf(' \n    New slip :\n    s_n = %f \n ',s_n3);\n",
+"printf(' \n    New rotor speed :\n    S_r = %f rpm \n',S_rn3);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.19: calculate_fcon_and_Scon.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-19\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"// Given data\n",
+"P = 8 ; // Number of poles in WRIM\n",
+"f = 60 ; // Operating frequency of the WRIM in Hz\n",
+"/// WRIM is driven by variable-speed prime mover as a frequency changer\n",
+"S_con_a1 = 1800 ; // Speed of the convertor in rpm\n",
+"S_con_a2 = 450 ; // Speed of the convertor in rpm\n",
+"f_con_b1 = 25 ; // Frequency of an induction converter in Hz\n",
+"f_con_b2 = 400 ; // Frequency of an induction converter in Hz\n",
+"f_con_b3 = 120 ; // Frequency of an induction converter in Hz\n",
+"// Calculations\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"// case a\n",
+"// Subscript a1 in f_con indicates case a 1st frequecy in Hz\n",
+"f_con_a1 = f*(1 + S_con_a1/S); // Frequency of an induction converter in Hz\n",
+"// Subscript a2 in f_con indicates case a 2nd frequency in Hz\n",
+"f_con_a2 = f*(1 - S_con_a2/S); // Frequency of an induction converter in Hz\n",
+"// case b\n",
+"// Subscript b1 in S-con indicates case b 1st speed of converter in rpm\n",
+"S_con_b1 = ( -1 + f_con_b1/f) * S ; // Speed of the convertor in rpm\n",
+"// Subscript b2 in S-con indicates case b 2nd speed of converter in rpm\n",
+"S_con_b2 = ( -1 + f_con_b2/f) * S ; // Speed of the convertor in rpm\n",
+"// Subscript b3 in S-con indicates case b 3rd speed of converter in rpm\n",
+"S_con_b3 = ( -1 + f_con_b3/f) * S ; // Speed of the convertor in rpm\n",
+"// Display the results\n",
+"disp('Example 9-19 Solution : ');\n",
+"printf(' \n Using Eq.(9-26),\n');\n",
+"printf(' \n a: f_con = %d Hz for %d rpm in opposite direction\n',f_con_a1,S_con_a1);\n",
+"printf(' \n    f_con = %d Hz for %d rpm in same direction\n',f_con_a2,S_con_a2);\n",
+"printf(' \n b: 1. S_con = %.f rpm, or %.f rpm in same direction.\n',S_con_b1,abs(S_con_b1));\n",
+"printf(' \n    2. S_con = %d rpm in opposite direction.\n',S_con_b2);\n",
+"printf(' \n    3. S_con = %d rpm in opposite direction to rotating stator flux.\n',S_con_b3);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.1: calculate_poles_and_synchronous_speed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-1\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"phase = 3 ; // Number of phases\n",
+"n = 3 ; // Slots per pole per phase\n",
+"f = 60 ; // Line frequency in Hz\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"P = 2 * n ; // Number of poles produced\n",
+"Total_slots = n * P * phase ; // Total number of slots on the stator\n",
+"\n",
+"// case b\n",
+"S_b = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"\n",
+"// case c\n",
+"f_c = 50 ; // Changed line frequency in Hz\n",
+"S_c = (120*f_c)/P ; // Speed in rpm of the rotating magnetic field\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-1 Solution : ');\n",
+"printf(' \n a: P = %d poles \n    Total slots = %d slots \n', P ,Total_slots );\n",
+"\n",
+"printf(' \n b: S = %d rpm  @ f = %d Hz \n ', S_b , f );\n",
+"\n",
+"printf(' \n c: S = %d rpm  @ f = %d Hz ', S_c ,f_c );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.2: calculate_rotor_speed.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-2\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"\n",
+"s_a = 5*(1/100); // Slip (case a)\n",
+"s_b = 7*(1/100); // Slip (case b)\n",
+"\n",
+"// Given data and calculated values from Ex.9-1\n",
+"f_a = 60 ; // Line frequency in Hz (case a)\n",
+"f_b = 50 ; // Line frequency in Hz (case b)\n",
+"S_a = 1200 ; // Speed in rpm of the rotating magnetic field (case a)\n",
+"S_b = 1000 ; // Speed in rpm of the rotating magnetic field (case b)\n",
+"\n",
+"// Calculations\n",
+"\n",
+"// case a\n",
+"S_r_a = S_a * ( 1 - s_a ); // Rotor speed in rpm when slip is 5% (case a)\n",
+"\n",
+"// case b\n",
+"S_r_b = S_b * ( 1 - s_b ); // Rotor speed in rpm when slip is 7% (case b)\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-2 Solution : ');\n",
+"\n",
+"printf(' \n a: S_r = %.f rpm  @ s = %.2f  \n ', S_r_a ,s_a );\n",
+"\n",
+"printf(' \n b: S_r = %.f rpm  @ s = %.2f  ', S_r_b ,s_b );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.3: calculate_rotor_frequency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-3\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 4 ; // Number of poles in Induction motor\n",
+"f = 60 ; // Frequency in Hz \n",
+"s_f = 5*(1/100) ; // Full-load rotor slip\n",
+"\n",
+"// Calculations\n",
+"\n",
+"// case a\n",
+"// slip, s = (S -S_r)/S ; \n",
+"// where S = Speed in rpm of the rotating magnetic field and \n",
+"//       S_r = Speed in rpm of the rotor\n",
+"s = 1 ; // Slip = 1, at the instant of starting, since S_r is zero\n",
+"f_r_a = s * f ; // Rotor frequency in Hz at the instant of starting\n",
+"\n",
+"// case b\n",
+"f_r_b = s_f * f ;// Full-load rotor frequency in Hz\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-3 Solution : ');\n",
+"\n",
+"printf(' \n a: At the instant of starting, slip s = (S -S_r)/S ; ');\n",
+"printf(' \n    where S_r is the rotor speed. Since the rotor speed at the ');\n",
+"printf(' \n    instant of starting is zero, s = (S - 0)/S = 1 , or unity slip.');\n",
+"printf(' \n\n    The rotor frequency is \n    f_r  = %d Hz \n\n ', f_r_a);\n",
+"\n",
+"printf(' \n b: At full-load,the slip is 5 percent(as given), and therefore');\n",
+"printf(' \n    s = %.2f \n    f_r = %d Hz ' , s_f , f_r_b);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.4: calculate_starting_torque_and_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-4\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 4 ; // Number of poles in the IM\n",
+"hp = 50 ; // rating of the IM in hp\n",
+"V_o = 208 ; // Voltage rating of the IM in volt\n",
+"T_orig = 225 ; // Starting torque in lb-ft\n",
+"I_orig = 700 ; // Instantaneous startign current in A at rated voltage\n",
+"V_s = 120 ; // Reduced 3-phase voltage supplied in volt\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"T_s = T_orig * (V_s/V_o)^2 ; // Starting torque in lb-ft after application of V_s\n",
+"\n",
+"// case b\n",
+"I_s = I_orig * (V_s/V_o) ; // Starting current in A after application of V_s\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-4 Solution : ');\n",
+"printf(' \n a: Starting torque :\n    T_s = %.f lb-ft \n',T_s );\n",
+"\n",
+"printf(' \n b: Starting current :\n    I_s = %d A \n',I_s );"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.5: calculate_s_Xlr_fr_Sr.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-5\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 8 ; // Number of poles in the SCIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"R_r = 0.3 ; // Rotor resistance per phase in ohm\n",
+"S_r = 650 ; // Speed in rpm at which motor stalls\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"s_b = (S - S_r)/S ; // Breakdown Slip\n",
+"\n",
+"// case b\n",
+"X_lr = R_r / s_b ; // Locked rotor reactance in ohm\n",
+"\n",
+"// case c\n",
+"f_r = s_b * f ; // Rotor frequency in Hz, at the maximum torque point \n",
+"\n",
+"// case d\n",
+"s = 5*(1/100);//  Rated slip\n",
+"S_r = S * (1 - s); // Full-load in rpm speed at rated slip \n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-5 Solution : ');\n",
+"printf(' \n a: S = %d rpm \n    s_b = %.3f \n', S , s_b );\n",
+"\n",
+"printf(' \n b: X_b = %.2f ohm \n ', X_lr );\n",
+"\n",
+"printf(' \n c: f_r = %.1f Hz \n ', f_r );\n",
+"\n",
+"printf(' \n d: S = %d rpm \n ', S_r );\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.6: calculate_full_load_S_and_Tf.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-6\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 8 ; // Number of poles in the SCIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"R_r = 0.3 ; // rotor resistance per phase in ohm/phase \n",
+"R_x = 0.7 ; // Added resistance in ohm/phase \n",
+"R_r_total = R_r + R_x ; // Total resistance per phase in ohm\n",
+"S_r = 875 ; // Full-load Speed in rpm \n",
+"\n",
+"\n",
+"// Calculated values from Ex.9-6\n",
+"S = 900 ; // Speed in rpm of the rotating magnetic field\n",
+"X_lr = 1.08 ; // Locked rotor reactance in ohm\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"s = (S - S_r)/S ; // Full-load slip,short circuited\n",
+"s_r = R_r_total / R_r * s; // New full-load slip with added resistance\n",
+"\n",
+"S_r_new = S*(1-s_r); // New full-load speed in rpm \n",
+"\n",
+"// case b\n",
+"// Neglecting constant Kn_t ,since we are taking torque ratios\n",
+"T_o = ( R_r / ((R_r)^2 + (X_lr)^2) ); // Original torque\n",
+"T_f = ( R_r + R_x) / ( (R_r + R_x)^2  + (X_lr)^2 ); // Original torque\n",
+"\n",
+"torque_ratio = T_f / T_o ; // Ratio of final torque to original torque\n",
+"T_final = 2*torque_ratio ;\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-6 Solution : ');\n",
+"printf(' \n a: The full-load slip,short circuited,is ');\n",
+"printf(' \n    s = %.4f \n',s );\n",
+"printf(' \n    Since slip is proportional to rotor resistance and since the ');\n",
+"printf(' \n    increased rotor resistance is R_r = %.1f + %.1f = %d ,',R_x,R_r,R_r_total);\n",
+"printf(' \n    the new full-load slip with added resistance is : ');\n",
+"printf(' \n    s_r = %.4f \n',s_r);\n",
+"printf(' \n    The new full-load speed is : ' );\n",
+"printf(' \n    S(1-s) = %.f rpm \n',S_r_new );\n",
+"\n",
+"printf(' \n b: The original starting torque T_o was twice the full-load torque');\n",
+"printf(' \n    with a rotor resistance of %.1f ohm and a rotor reactance of %.2f ohm',R_r,X_lr);\n",
+"printf(' \n    (Ex.9-5).The new starting torque conditions may be summarized by the   ');\n",
+"printf(' \n    following table and compared from Eq.(9-14),where T_o ');\n",
+"printf(' \n    is the original torque and T_f is the new torque.');\n",
+"\n",
+"printf(' \n    _________________________________________');\n",
+"printf(' \n    Condition \t R_r \t X_lr \t T_starting ');\n",
+"printf(' \n              \t ohm \t ohm  \t ');\n",
+"printf(' \n    _________________________________________');\n",
+"printf(' \n    Original : \t %.1f \t %.2f \t 2*T_n ',R_r,X_lr);\n",
+"printf(' \n    New      : \t %.1f \t %.2f \t   ?   ',R_r_total,X_lr);\n",
+"printf(' \n    _________________________________________\n');\n",
+"\n",
+"printf(' \n    T_o = %.2f * K_n_t',T_o);\n",
+"printf(' \n    T_f = %.3f * K_n_t',T_f);\n",
+"printf(' \n    T_f/T_o = %.2f and T_f = %.2f * T_o\n ',torque_ratio,torque_ratio);\n",
+"printf(' \n    Therefore,\n    T_f = %.3f * T_n',T_final);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.7: calculate_rotor_I_and_PF_and_same_with_added_Rr.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-7\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 8 ; // Number of poles in the SCIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"R_r = 0.3 ; // Rotor resistance per phase in ohm\n",
+"R_x = 0.7 ; // Added resistance in ohm/phase \n",
+"R_r_total = R_r + R_x ; // Total resistance per phase in ohm\n",
+"X_lr = 1.08 ; // Locked rotor reactance in ohm\n",
+"S_r = 650 ; // Speed in rpm at which motor stalls\n",
+"E_lr = 112 ; // Induced voltage per phase\n",
+"\n",
+"// Calculations\n",
+"// case a\n",
+"Z_lr = R_r + %i*X_lr ; // Locked rotor impedance per phase\n",
+"Z_lr_m = abs(Z_lr);//Z_lr_m = magnitude of Z_lr in ohm\n",
+"Z_lr_a = atan(imag(Z_lr) /real(Z_lr))*180/%pi;//Z_lr_a=phase angle of Z_lr in degrees\n",
+"\n",
+"I_r = E_lr / Z_lr_m ; // Rotor current per phase\n",
+"cos_theta_r = cosd(Z_lr_a); // rotor power factor with the rotor short-circuited\n",
+"cos_theta = R_r / Z_lr_m ; // rotor power factor with the rotor short-circuited\n",
+"\n",
+"// case b\n",
+"// 1 at the end of Z_lr1 is just used for showing its different form Z_lr \n",
+"// and for ease in calculations\n",
+"Z_lr1 = R_r_total + %i*X_lr ; // Locked rotor impedance per phase\n",
+"Z_lr1_m = abs(Z_lr1);//Z_lr1_m = magnitude of Z_lr1 in ohm\n",
+"Z_lr1_a = atan(imag(Z_lr1) /real(Z_lr1))*180/%pi;//Z_lr1_a=phase angle of Z_lr1 in degrees\n",
+"\n",
+"I_r1 = E_lr / Z_lr1_m ; // Rotor current per phase\n",
+"cos_theta_r1 = cosd(Z_lr1_a); // rotor power factor with the rotor short-circuited\n",
+"cos_theta1 = R_r_total / Z_lr1_m ; // rotor power factor with the rotor short-circuited\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-7 Solution : ');\n",
+"printf(' \n a: The locked-rotor impedance per phase is : ');\n",
+"printf(' \n    Z_lr in ohm = '),disp(Z_lr);\n",
+"printf(' \n    Z_lr = %.2f <%.1f ohm \n',Z_lr_m,Z_lr_a);\n",
+"printf(' \n    I_r = %.f A \n',I_r);\n",
+"printf(' \n    cosθ_r = cos(%.1f) = %.3f or \n    cosθ = R_r/Z_lr = %.3f',Z_lr_a,cos_theta_r,cos_theta);\n",
+"\n",
+"printf(' \n\n\n b: The locked-rotor impedance with added rotor resistance per phase is : ');\n",
+"printf(' \n    Z_lr in ohm = '),disp(Z_lr1);\n",
+"printf(' \n    Z_lr = %.2f <%.1f ohm \n',Z_lr1_m,Z_lr1_a);\n",
+"printf(' \n    I_r = %.1f A \n',I_r1);\n",
+"printf(' \n    cosθ_r = cos(%.1f) = %.3f or \n    cosθ = R_r/Z_lr = %.3f',Z_lr1_a,cos_theta_r1,cos_theta1);"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.8: calculate_Rx_and_rotor_PF_and_starting_current.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-8\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data  (Exs.9-5 through 9-7)\n",
+"P = 8 ; // Number of poles in the SCIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"R_r = 0.3 ; // Rotor resistance per phase in ohm\n",
+"X_lr = 1.08 ; // Locked rotor reactance in ohm\n",
+"S_r = 650 ; // Speed in rpm at which motor stalls\n",
+"E_lr = 112 ; // Induced voltage per phase\n",
+"\n",
+"disp('Example 9-8 : '); \n",
+"printf(' \n The new and the original conditions may be summarized in the following table\n');\n",
+"printf(' \n    _________________________________________________________');\n",
+"printf(' \n    Condition \t R_r \t\t X_lr \t\t T_starting ');\n",
+"printf(' \n              \t ohm \t\t ohm  \t ');\n",
+"printf(' \n    _________________________________________________________');\n",
+"printf(' \n    Original : \t %.1f \t\t %.2f \t\t T_o = 2*T_n ',R_r,X_lr);\n",
+"printf(' \n    New      :\t(%.1f+R_x) \t %.2f \t\t T_n = 2*T_n ',R_r,X_lr);\n",
+"printf(' \n    _________________________________________________________\n');\n",
+"\n",
+"// Calculating\n",
+"// case a\n",
+"// Neglecting constant Kn_t ,since we are equating torque T_o and T_n\n",
+"T_o = ( R_r / ((R_r)^2 + (X_lr)^2) ); // Original torque\n",
+"\n",
+"// T_o = K_n_t*( 0.3 / ((0.3)^2 + (1.08)^2) );\n",
+"// T_n = K_n_t*( 0.3 + R_x) / ( (0.3 + R_x)^2  + (1.08)^2 );\n",
+"// T_n = T_o\n",
+"// Simplyifing yields\n",
+"// 0.3 + R_x = 0.24[(0.3+R_x)^2 + (1.08)^2]\n",
+"// Expanding and combining the terms yields\n",
+"// 0.24*(R_x)^2 - 0.856*R_x = 0\n",
+"// This is a quadratic equation having two roots,which may be factored as\n",
+"// R_x*(0.24*R_x - 0.856) = 0,yielding\n",
+"// R_x = 0 and R_x = 0.856/0,24 = 3.57\n",
+"R_x = poly(0,'R_x'); // Defining a polynomial with variable 'R_x' with  root at 0\n",
+"a = 0.24 ; // coefficient of x^2\n",
+"b = -0.856 ; // coefficient of x\n",
+"c = 0 ; // constant\n",
+"\n",
+"// Roots of p\n",
+"R_x1 = ( -b + sqrt (b^2 -4*a*c ) ) /(2* a);\n",
+"R_x2=( -b - sqrt (b^2 -4*a*c ) ) /(2* a);\n",
+"// Consider R_x>0 value, \n",
+"R_x = R_x1;\n",
+"\n",
+"R_T = R_r + R_x ; // Total rotor resistance in ohm\n",
+"\n",
+"// case b\n",
+"Z_T = R_T + %i*X_lr ; // Total impedance in ohm\n",
+"Z_T_m = abs(Z_T);//Z_T_m = magnitude of Z_T in ohm\n",
+"Z_T_a = atan(imag(Z_T) /real(Z_T))*180/%pi;//Z_T_a=phase angle of Z_T in degrees\n",
+"\n",
+"cos_theta = R_T / Z_T_m ; // Rotor PF that will produce the same starting torque\n",
+"\n",
+"// case c\n",
+"Z_r = Z_T_m ; // Impedance in ohm\n",
+"I_r = E_lr / Z_r ; // Starting current in A\n",
+"\n",
+"// Display the results\n",
+"disp('Solution : '); \n",
+"\n",
+"printf(' \n a: T_o = %.2f * K_n_t ',T_o );\n",
+"printf(' \n    T_n = %.2f * K_n_t \n',T_o );\n",
+"printf(' \n    Simplyifing yields');\n",
+"printf(' \n    0.3 + R_x = 0.24[(0.3+R_x)^2 + (1.08)^2]');\n",
+"printf(' \n    Expanding and combining the terms yields');\n",
+"printf(' \n    0.24*(R_x)^2 - 0.856*R_x = 0');\n",
+"printf(' \n    This is a quadratic equation having two roots,which may be factored as');\n",
+"printf(' \n    R_x*(0.24*R_x - 0.856) = 0,yielding');\n",
+"printf(' \n    R_x = 0 ohm and R_x = 0.856/0.24 = 3.57 ohm\n\n    This proves that ');\n",
+"printf(' \n    Original torque is produced with an external resistance of either ');\n",
+"printf(' \n    zero or 12 times the origianl rotor resistance.Therefore,\n');\n",
+"printf(' \n    R_T = R_r + R_x = %.2f ohm \n',R_T);\n",
+"\n",
+"printf(' \n b: Z_T in ohm = ');disp(Z_T);\n",
+"printf(' \n    Z_T = %.2f <%.1f ohm ',Z_T_m,Z_T_a);\n",
+"printf(' \n    cosӨ = R_T / Z_T = %.3f or \n    cosӨ = cosd(%.1f) = %.3f\n',cos_theta,Z_T_a,cosd(Z_T_a));\n",
+"\n",
+"printf(' \n c: I_r = E_lr / Z_r = %.f A \n\n    This proves that,',I_r);\n",
+"printf(' \n    Rotor current at starting is now only 28 percent of the original');\n",
+"printf(' \n    starting current in part(a) of Ex.9-7');"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 9.9: calculate_Sr_with_added_Rx.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Electric Machinery and Transformers\n",
+"// Irving L kosow \n",
+"// Prentice Hall of India\n",
+"// 2nd editiom\n",
+"\n",
+"// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS\n",
+"// Example 9-9\n",
+"\n",
+"clear; clc; close; // Clear the work space and console.\n",
+"\n",
+"// Given data\n",
+"P = 8 ; // Number of poles in the SCIM\n",
+"f = 60 ; // Frequency in Hz\n",
+"S_r = 875 ; // Full-load Speed in rpm with rotor short-circuited\n",
+"R_r = 0.3 ; // rotor resistance per phase in ohm/phase \n",
+"R_x = 0.7 ; // Added resistance in ohm/phase \n",
+"R_x_a = 1.7 ; // Added resistance in ohm/phase (case a)\n",
+"R_x_b = 2.7 ; // Added resistance in ohm/phase (case b) \n",
+"R_x_c = 3.7 ; // Added resistance in ohm/phase (case c) \n",
+"R_x_d = 4.7 ; // Added resistance in ohm/phase (case d) \n",
+"\n",
+"// Calculations\n",
+"S = (120*f)/P ; // Speed in rpm of the rotating magnetic field\n",
+"s_o = (S - S_r)/S ; //  Slip at rotor speed 875 rpm\n",
+"\n",
+"// case a\n",
+"s_r_a = s_o * (R_r + R_x_a)/R_r; // Rated slip\n",
+"S_r_a = S * (1 - s_r_a); // Full-load speed in rpm for added resistance R_x_a\n",
+"\n",
+"// case b\n",
+"s_r_b = s_o * (R_r + R_x_b)/R_r; // Rated slip\n",
+"S_r_b = S * (1 - s_r_b); // Full-load speed in rpm for added resistance R_x_b\n",
+"\n",
+"// case c\n",
+"s_r_c = s_o * (R_r + R_x_c)/R_r; // Rated slip\n",
+"S_r_c = S * (1 - s_r_c); // Full-load speed in rpm for added resistance R_x_c\n",
+"\n",
+"// case d\n",
+"s_r_d = s_o * (R_r + R_x_d)/R_r; // Rated slip\n",
+"S_r_d = S * (1 - s_r_d); // Full-load speed in rpm for added resistance R_x_d\n",
+"\n",
+"// Display the results\n",
+"disp('Example 9-9 Solution : ');\n",
+"\n",
+"printf(' \n Slip s_r = s_o*(R_r+R_x)/R_r \n Rotor speed S_r = S_o*(1-s)\n');\n",
+"\n",
+"printf(' \n    Calculated value of s_o = %f , instead of 0.0278(textbook)',s_o)\n",
+"printf(' \n    so slight variations in the answers below.\n');\n",
+"\n",
+"printf(' \n a: When R_x = %.1f ohm ',R_x_a);\n",
+"printf(' \n    s_r = %.3f \n    S_r = %.1f rpm \n',s_r_a,S_r_a );\n",
+"\n",
+"printf(' \n b: When R_x = %.1f ohm ',R_x_b);\n",
+"printf(' \n    s_r = %.3f \n    S_r = %.1f rpm \n',s_r_b,S_r_b );\n",
+"\n",
+"printf(' \n c: When R_x = %.1f ohm ',R_x_c);\n",
+"printf(' \n    s_r = %.3f \n    S_r = %.1f rpm \n',s_r_c,S_r_c );\n",
+"\n",
+"printf(' \n d: When R_x = %.1f ohm ',R_x_d);\n",
+"printf(' \n    s_r = %.3f \n    S_r = %.1f rpm \n',s_r_d,S_r_d );\n",
+"\n",
+"printf(' \n    This example,verifies that slip is proportional to rotor resistance');\n",
+"printf(' \n    as summarized below.');\n",
+"\n",
+"printf(' \n  ___________________________________________________________________');\n",
+"printf(' \n    R_T(ohm) = R_r+R_x \t\t Slip \t\t Full-load Speed(rpm)');\n",
+"printf(' \n  ___________________________________________________________________');\n",
+"printf(' \n      Given \t\t\t Given \t\t Given \t\ ');\n",
+"printf(' \n      0.3 \t\t\t 0.0278 \t 875 ');\n",
+"printf(' \n      0.3+0.1 = 1.0 \t\t 0.0926 \t 817');\n",
+"printf(' \n  ___________________________________________________________________');\n",
+"printf(' \n      Given \t\t\t Calculated \t Calculated \t\ ');\n",
+"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);\n",
+"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);\n",
+"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);\n",
+"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);\n",
+"printf(' \n  ___________________________________________________________________');"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}