Added(A)/Deleted(D) following books

A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/README.txt A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch2_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch3_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch4_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch5_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch6_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch7_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch8_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch9_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/ch2_1.png A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/ch7_1.png A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/kVSv5.png A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER2_1.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER4_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER5_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER6_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER7_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/README.txt A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/screenshots/CHAP2.png A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/screenshots/CHAP4.png A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/screenshots/CHAP5.png D A_First_course_in_Programming_with_C/Chapter14.ipynb M A_First_course_in_Programming_with_C_by_T_Jeyapoovan/Chapter14_2.ipynb A A_First_course_in_Programming_with_C_by_T_Jeyapoovan/README.txt A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1.2.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1.3.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1.7.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1_2.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1_3.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1_7.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.10.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.11.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.13.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.14.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.15.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.16.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.17.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.18.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.2.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.3.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.4.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.5.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.6.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.7.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.8.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.9.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2_10.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2_11.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2_13.ipynb A 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Internal_Combustion_Engines_by_H._B._Keswani/screenshots/ch9.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter12_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter12_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter12_3.ipynb A 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Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/README.txt A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_1.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_2.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_3.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_4.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_1.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_2.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_3.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_4.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_1.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_2.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_3.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_4.png A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter1.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter2(PartB).ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter2.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter3(partB).ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter3.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter4(PartB).ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter4.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter5.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter6.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter7.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/screenshots/chapter1.png A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/screenshots/chapter2.png A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/screenshots/chapter3.png A Introduction_to_Electric_Drives_by_J._S._Katre/AppendixB.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/AppendixB_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter10.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter10_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter10_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter1_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter1_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter2_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter2_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter3.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter3_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter3_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter5.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter5_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter5_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter6.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter6_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter6_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter8.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter8_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter8_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter9.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter9_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter9_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_VLdc_VLrms.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_VLdc_VLrms_1.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_VLdc_VLrms_2.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_variation_of_RF_FF.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_variation_of_RF_FF_1.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_variation_of_RF_FF_2.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch_3_variation_avg_rms_load_V.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch_3_variation_avg_rms_load_V_1.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch_3_variation_avg_rms_load_V_2.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex1.2.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex3.13.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex6.7.png A Linear_Integrated_Circuits_by_J._B._Gupta/README.txt A Linear_Integrated_Circuits_by_J._B._Gupta/chapter01_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter01_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter02_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter02_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter03_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter03_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter04_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter04_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter05_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter05_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter06_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter06_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter07_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter07_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter08_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter08_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter09_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter09_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter10_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter10_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter11_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter11_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_14.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_14_1.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_15.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_15_1.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/per_error_1.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/per_error_2.png A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER19.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER21.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER24_.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER26.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER30.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER31.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER33.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP10.png A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP16.png A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP19.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/FricCoeff.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/FricCoeff_1.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/fillingtime.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/fillingtime_1.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/millPOwer.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/millPOwer_1.png A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter10.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter11.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter12.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter13.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter14.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter15.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter16.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter17.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter2.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter3.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter4.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter5.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter6.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter7.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter8.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter9.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/screenshots/Chapter10.png A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/screenshots/Chapter11.png A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/screenshots/Chapter12.png A Materials_Science_by_Dr._M._Arumugam/Chapter10_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter12_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter1_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter2_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter3_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter4_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter5_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter6_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter7_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter8_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter9_1.ipynb A Materials_Science_by_Dr._M._Arumugam/README.txt A Materials_Science_by_Dr._M._Arumugam/screenshots/11.png A Materials_Science_by_Dr._M._Arumugam/screenshots/22.png A Materials_Science_by_Dr._M._Arumugam/screenshots/33.png A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_2_Generalized_Configurations_and_Functional_Descriptions_of_Measuring_Instruments.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_3_Generalized_Performance_Characteristics_of_Instruments.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_4_Relative_Velocity_Translational_and_Rotational.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_5_Force_Torque_and_Shaft_power_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_6_Pressure_and_Sound_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_7_Flow_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_8_Temperature_and_Heat-Flux_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/screenshots/cha3.png A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/screenshots/cha4.png A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/screenshots/cha5.png A Mechanical_Metallurgy_by_George_E._Dieter/README.txt A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_10.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_11.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_12.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_13.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_14.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_4.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_5.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_6.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_7.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_8.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_9.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_3.ipynb A 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Special_Electrical_Machines_by_S.P._Burman/chapter03.ipynb A Special_Electrical_Machines_by_S.P._Burman/chapter04.ipynb A Special_Electrical_Machines_by_S.P._Burman/screenshots/ResolShaftSpeed3.png A Special_Electrical_Machines_by_S.P._Burman/screenshots/TorqLossEff1.png A Special_Electrical_Machines_by_S.P._Burman/screenshots/Torq_Speed1.png A Strength_Of_Materials_by_B_K_Sarkar/Chapter01.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter02.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter03.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter04.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter05.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter06.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter07.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter08.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter09.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter10.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter11.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter12.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter13.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter14.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter15.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter16.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter17.ipynb A Strength_Of_Materials_by_B_K_Sarkar/README.txt A Strength_Of_Materials_by_B_K_Sarkar/chapter_10_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_10_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_11_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_11_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_12_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_12_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_13_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_13_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_14_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_14_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_15_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_15_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_16_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_16_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_17_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_17_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_1_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_1_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_2_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_2_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_3_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_3_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_4_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_4_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_5_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_5_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_6_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_6_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_7_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_7_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_8_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_8_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_9_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_9_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/screenshots/B.M.D_1.JPG A Strength_Of_Materials_by_B_K_Sarkar/screenshots/B.M.D_2.JPG A Strength_Of_Materials_by_B_K_Sarkar/screenshots/BMD2.png A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_1.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_1_1.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_2.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_4.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/SFD.png A Strength_Of_Materials_by_B_K_Sarkar/screenshots/SFD3.png M The_C_Book/Chapter2.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/README.txt R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch10.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch10.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch10_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch2.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch2.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch2_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch3.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch3.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch3_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch4.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch4.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch4_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch5.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch5.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch5_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch6.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch6.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch6_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch7.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch7.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch7_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch8.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch8.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch8_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch9.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch9.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch9_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same3_7.png -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same3_7.png A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same3_7_1.png R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same7.png -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same7.png A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same7_1.png R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/shearAndBendingMoment7.png -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/shearAndBendingMoment7.png A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/shearAndBendingMoment7_1.png A Wireless_Communications_and_Networking_by_V._Garg/README.txt A Wireless_Communications_and_Networking_by_V._Garg/ch10_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch11_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch12_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch13_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch14_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch17_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch19_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch21_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch2_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch3_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch4_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch5_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch6_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch8_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch9_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/screenshots/EbbyNo_1.png A Wireless_Communications_and_Networking_by_V._Garg/screenshots/comparision_of_models_1.png A Wireless_Communications_and_Networking_by_V._Garg/screenshots/multiplexing_1.png A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/README.txt A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch10_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch11_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch12_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch1_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch2_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch3_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch4_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch5_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch6_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch7_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch8_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch9_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/screenshots/energy_stored3_1.png A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/screenshots/magnetic_flux12_1.png A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/screenshots/pitch_factor7_1.png A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium_2.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium_3.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams_2.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams_3.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment_2.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment_3.ipynb A 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author: FOSSEE SysAds 2015-12-08 15:04:13 +0600
committer: FOSSEE SysAds 2015-12-08 15:04:13 +0600
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diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/README.txt b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/README.txt
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+Contributed By: Paritosh Mehta
+Course: me
+College/Institute/Organization: GTU
+Department/Designation: VLSI
+Book Title: Electric Power Distribution System Engineering
+Author: T. Gonen
+Publisher: McGraw Hill, New York
+Year of publication: 1986
+Isbn: 0-07-023707-7
+Edition: 1
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch10_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch10_1.ipynb
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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bc905e996740356ca2e367534c643c16115b03b37f0c3f1fca3b88fd2929cfdd"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Distribution System Protection"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1  Page No : 542"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "#For Recloser\n",
+      "InstT = 0.03;         #From Curve A         #instaneous Time\n",
+      "TimeD = 0.17;         #From Curve B         #Time Delay\n",
+      "#For Relay\n",
+      "PickU = 0.42;         #From Curve C         #Pick Up \n",
+      "Reset = (1./10)*60;         #Assuming a 60 s reset time for the relay with number 10 time dial setting\n",
+      "RecloserOT = 1;         #Assumed Recloser Open Time\n",
+      "\n",
+      "RelayCTI = InstT/PickU;         #Relay Closing Travel during instanmath.taneous operation\n",
+      "RelayRTI = (-1)*RecloserOT/Reset;         #Relay Reset Travel during instanmath.taneuos\n",
+      "\n",
+      "# Calculations\n",
+      "RelayCTD = TimeD/PickU;\n",
+      "RelayRTD = (-1)*RecloserOT/Reset;         #Relay Reset Travel during trip\n",
+      "NetRelayTravel = RelayCTD-RelayRTD;\n",
+      "\n",
+      "# Results\n",
+      "print 'During instanmath.taneous Operation'\n",
+      "print '|Relay Closing Travel| < |Relay Rest Travel|'\n",
+      "print '|%g percent| < |%g percent|'%(RelayCTI*100,RelayRTI*100)\n",
+      "\n",
+      "print 'During the Delayed Tripping Operations'\n",
+      "print 'Total Relay Travel is from:'\n",
+      "print '%g percent to %g percent to %g percent'%(RelayCTD*100,RelayRTD*100,RelayCTD*100)\n",
+      "print 'Since this Net Total Relay Travel is less than 100 percent the desired recloser to relay coordination is accomplished'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "During instanmath.taneous Operation\n",
+        "|Relay Closing Travel| < |Relay Rest Travel|\n",
+        "|7.14286 percent| < |-16.6667 percent|\n",
+        "During the Delayed Tripping Operations\n",
+        "Total Relay Travel is from:\n",
+        "40.4762 percent to -16.6667 percent to 40.4762 percent\n",
+        "Since this Net Total Relay Travel is less than 100 percent the desired recloser to relay coordination is accomplished\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2  Page No : 555"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Vln = 7200.;         #Line to Neutral Voltage\n",
+      "Vll = 12470.;         #Line to Line Voltage\n",
+      "Z1sys = 0.7199+(1j*3.4619);         #system impedance to the regulated 12.47kV bus\n",
+      "ZGsys = 0.6191+(1j*3.3397);        #system impedance to ground\n",
+      "l = 2;         #Dismath.tance of the Fault from the substation\n",
+      "        #From Table 10-7; Various Paramters Can Be found out\n",
+      "z0a = 0.1122+(1j*0.4789);\n",
+      "z011 = (-0.0385-(1j*0.0996));\n",
+      "z1 = 0.0580+(1j*0.1208);\n",
+      "C = 5.28;         #Cable consmath.tant\n",
+      "\n",
+      "# Calculations\n",
+      "Z0ckt = 2*(z0a+z011)*C;         #Zero Sequence Impedance\n",
+      "Z1ckt = 2*z1*C;         #Positive Sequence Impedance\n",
+      "ZGckt = ((2*Z1ckt)+Z0ckt)/3;         #Impedance to ground of line\n",
+      "        #Note That the calculation of the above term is wrong in the text book\n",
+      "\n",
+      "Z1 = Z1sys+Z1ckt;         #Total Positive Sequence\n",
+      "ZG = ZGsys+ZGckt;         #Total impedance to ground\n",
+      "\n",
+      "If3phi = Vln/abs(Z1);         #Three Phase Fault at point 10\n",
+      "IfLL = 0.866*If3phi;         #Line to Line Fault at point 10\n",
+      "IfLG = Vln/(abs(ZG));         #Single Line to Ground Fault\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Zero and Postive sequence impedance of the line to point 10 are:',\n",
+      "print (Z0ckt),\n",
+      "print (Z1ckt)\n",
+      "print 'b The impedance to ground of the line to point 10',\n",
+      "print (ZGckt)\n",
+      "print 'c The Total positive sequence impedance including system impedance is',\n",
+      "print (Z1)\n",
+      "print 'd The Total Impedance to ground to point 10 including system impedance is',\n",
+      "print (ZG)\n",
+      "print 'All the Above impedances are in ohm'\n",
+      "print 'e) The Three phase fault current at point 10 is %g A'%(If3phi)\n",
+      "print 'f) The line to line fault current at point 10 is %g A'%(IfLL)\n",
+      "print 'g) The Line to Ground Current at point 10 is %g A'%(IfLG)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Zero and Postive sequence impedance of the line to point 10 are: (0.778272+4.005408j) (0.61248+1.275648j)\n",
+        "b The impedance to ground of the line to point 10 (0.667744+2.185568j)\n",
+        "c The Total positive sequence impedance including system impedance is (1.33238+4.737548j)\n",
+        "d The Total Impedance to ground to point 10 including system impedance is (1.286844+5.525268j)\n",
+        "All the Above impedances are in ohm\n",
+        "e) The Three phase fault current at point 10 is 1463.02 A\n",
+        "f) The line to line fault current at point 10 is 1266.97 A\n",
+        "g) The Line to Ground Current at point 10 is 1269.14 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3  Page No : 562"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "St = 5*(10**6);         #Capacity of Transformer\n",
+      "Zt = 1j*0.065;         #Transformer Reacmath.tance\n",
+      "SB3phi = 1*(10**6);         #3 Phase Power Base\n",
+      "VBLL = 69*(10**3);         #Line to line voltage\n",
+      "VBLLn = 12.47*(10**3);         #Line To line voltage\n",
+      "Vf = 1;         #Per Unit Value of Voltage\n",
+      "Zb = (VBLL**2)/SB3phi;         #Base Impedance\n",
+      "\n",
+      "#Zckt and Zf and Zt are Zero for Bus 1\n",
+      "#Zckt and Zf are Zero for Bus 2\n",
+      "#Power Generation of the system\n",
+      "SMax = 600*(10**6);         #Maximum\n",
+      "SMin = 360*(10**6);         #Minimum\n",
+      "\n",
+      "# Calculations\n",
+      "Xt = 0.065;         #Transformer Reacmath.tance in per unit\n",
+      "MZsysmax = (VBLL**2)/SMax;         #System Impedance at Maximum Power Generation\n",
+      "Ib = SB3phi/(math.sqrt(3)*VBLL);         #Base Current\n",
+      "Zsysmaxpu = MZsysmax*exp(1j*math.pi*90/180)/Zb;         #System Impedance Phasor\n",
+      "#Three Phase Fault Current\n",
+      "If3phimaxpu1 = abs(Vf/(Zsysmaxpu));\n",
+      "If3phimax1 = If3phimaxpu1*Ib;\n",
+      "Sf3phimax1 = math.sqrt(3)*VBLL*If3phimax1/1000000;\n",
+      "\n",
+      "#Line to Line Fault Current\n",
+      "IfLLmax1 = 0.866*If3phimax1;\n",
+      "SfLLmax1 = VBLL*IfLLmax1/1000000;\n",
+      "\n",
+      "#Line to Ground Fault\n",
+      "IfLGmaxpu1 = abs(3*Vf/((2*Zsysmaxpu)));\n",
+      "IfLGmax1 = IfLGmaxpu1*Ib;\n",
+      "SfLGmax1 = VBLL*IfLGmax1/(1000000*math.sqrt(3));\n",
+      "\n",
+      "Stn = SB3phi;         #Numreical Value is Equal\n",
+      "Ztn = Zt*(Stn/St);         #New Per Unit Transformer Reacmath.tance\n",
+      "#New Base Values\n",
+      "Zbn = (VBLLn**2)/SB3phi;\n",
+      "Ibn = Stn/(math.sqrt(3)*VBLLn);\n",
+      "\n",
+      "#Three Phase Fault Current\n",
+      "If3phimaxpu2 = abs(Vf/(Zsysmaxpu+Ztn));\n",
+      "If3phimax2 = If3phimaxpu2*Ibn;\n",
+      "Sf3phimax2 = math.sqrt(3)*VBLLn*If3phimax2/1000000;\n",
+      "\n",
+      "#Line to Line Fault Current\n",
+      "IfLLmax2 = 0.866*If3phimax2;\n",
+      "SfLLmax2 = VBLLn*IfLLmax2/1000000;\n",
+      "\n",
+      "#Line to Ground Fault\n",
+      "IfLGmaxpu2 = abs(3*Vf/((2*Zsysmaxpu)+(3*Ztn)));\n",
+      "IfLGmax2 = IfLGmaxpu2*Ibn;\n",
+      "SfLGmax2 = VBLLn*IfLGmax2/(1000000*math.sqrt(3));\n",
+      "\n",
+      "#Minimum Power Generation\n",
+      "MZsysmin = (VBLL**2)/SMin;         #System Impedance at Maximum Power Generation\n",
+      "Ib = SB3phi/(math.sqrt(3)*VBLL);         #Base Current\n",
+      "Zsysminpu = MZsysmin*exp(1j*math.pi*90/180)/Zb;         #System Impedance Phasor\n",
+      "#Three Phase Fault Current\n",
+      "If3phiminpu1 = abs(Vf/(Zsysminpu));\n",
+      "If3phimin1 = If3phiminpu1*Ib;\n",
+      "Sf3phimin1 = math.sqrt(3)*VBLL*If3phimin1/1000000;\n",
+      "\n",
+      "#Line to Line Fault Current\n",
+      "IfLLmin1 = 0.866*If3phimin1;\n",
+      "SfLLmin1 = VBLL*IfLLmin1/1000000;\n",
+      "\n",
+      "#Line to Ground Fault\n",
+      "IfLGminpu1 = abs(3*Vf/((2*Zsysminpu)));\n",
+      "IfLGmin1 = IfLGminpu1*Ib;\n",
+      "SfLGmin1 = VBLL*IfLGmin1/(1000000*math.sqrt(3));\n",
+      "\n",
+      "Stn = SB3phi;         #Numreical Value is Equal\n",
+      "Ztn = Zt*(Stn/St);         #New Per Unit Transformer Reacmath.tance\n",
+      "#New Base Values\n",
+      "Zbn = (VBLLn**2)/SB3phi;\n",
+      "Ibn = Stn/(math.sqrt(3)*VBLLn);\n",
+      "\n",
+      "#Three Phase Fault Current\n",
+      "If3phiminpu2 = abs(Vf/(Zsysminpu+Ztn));\n",
+      "If3phimin2 = If3phiminpu2*Ibn;\n",
+      "Sf3phimin2 = math.sqrt(3)*VBLLn*If3phimin2/1000000;\n",
+      "\n",
+      "#Line to Line Fault Current\n",
+      "IfLLmin2 = 0.866*If3phimin2;\n",
+      "SfLLmin2 = VBLLn*IfLLmin2/1000000;\n",
+      "\n",
+      "#Line to Ground Fault\n",
+      "IfLGminpu2 = abs(3*Vf/((2*Zsysminpu)+(3*Ztn)));\n",
+      "IfLGmin2 = IfLGminpu2*Ibn;\n",
+      "SfLGmin2 = VBLLn*IfLGmin2/(1000000*math.sqrt(3));\n",
+      "\n",
+      "# Results\n",
+      "print 'a For Maximum Power Generation:'\n",
+      "print 'Bus 1'\n",
+      "print '3 phase fault current is %g A and %g MVA'%(If3phimax1,Sf3phimax1)\n",
+      "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmax1,SfLLmax1)\n",
+      "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmax1,SfLGmax1)\n",
+      "print 'Bus 2'\n",
+      "print '3 phase fault current is %g A and %g MVA'%(If3phimax2,Sf3phimax2)\n",
+      "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmax2,SfLLmax2)\n",
+      "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmax2,SfLGmax2)\n",
+      "print 'b For Minimum Power Generation:'\n",
+      "print 'Bus 1'\n",
+      "print '3 phase fault current is %g A and %g MVA'%(If3phimin1,Sf3phimin1)\n",
+      "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmin1,SfLLmin1)\n",
+      "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmin1,SfLGmin1)\n",
+      "print 'Bus 2'\n",
+      "print '3 phase fault current is %g A and %g MVA'%(If3phimin2,Sf3phimin2)\n",
+      "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmin2,SfLLmin2)\n",
+      "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmin2,SfLGmin2)\n",
+      "\n",
+      "#Note that 0.00166666666 is not rounded as 0.0017\n",
+      "#Hence you find all the answers close by\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a For Maximum Power Generation:\n",
+        "Bus 1\n",
+        "3 phase fault current is 5691.02 A and 680.143 MVA\n",
+        "Line to Line fault current is 4928.43 A and 340.061 MVA\n",
+        "Line to ground fault current is 8536.54 A and 340.071 MVA\n",
+        "Bus 2\n",
+        "3 phase fault current is 31490 A and 680.143 MVA\n",
+        "Line to Line fault current is 27270.4 A and 340.061 MVA\n",
+        "Line to ground fault current is 47235 A and 340.071 MVA\n",
+        "b For Minimum Power Generation:\n",
+        "Bus 1\n",
+        "3 phase fault current is 3064.4 A and 366.231 MVA\n",
+        "Line to Line fault current is 2653.77 A and 183.11 MVA\n",
+        "Line to ground fault current is 4596.6 A and 183.115 MVA\n",
+        "Bus 2\n",
+        "3 phase fault current is 16956.2 A and 366.231 MVA\n",
+        "Line to Line fault current is 14684 A and 183.11 MVA\n",
+        "Line to ground fault current is 25434.3 A and 183.115 MVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4  Page No : 572"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Percent Impedances of the substation transformer\n",
+      "Rtp = 1.;\n",
+      "Ztp = 7.;\n",
+      "Xtp = math.sqrt((Ztp**2)-(Rtp**2)); \n",
+      "Ztpu = Rtp+(1j*Xtp);         #Transformer Impedance\n",
+      "Vll = 12.47;         #Line to Line voltage in kV\n",
+      "Vln = 7.2;         #Line to Neutral Voltage\n",
+      "V = 240.;         #Secondary Voltage\n",
+      "St = 7500.;         #Rating of the transformer in kVA\n",
+      "Sts = 100.;         #Rating of Secondary Transformer\n",
+      "Ztp = Ztpu*((Vll**2)*10/St);\n",
+      "SSC = complex(.466,0.0293);\n",
+      "#From Table 10-7\n",
+      "Z1 = 0.0870+(1j*0.1812);\n",
+      "Z0 = complex(0.1653,0.4878);\n",
+      "\n",
+      "ZG = ((2*Z1)+Z0)/3;        #Impedance to Ground\n",
+      "\n",
+      "Zsys = 0 ;         #Assumption Made\n",
+      "Zeq = Zsys+Ztp+ZG;         #Equivalent Impedance of the Primary\n",
+      "\n",
+      "PZ2 = Zeq*((V/(Vln*1000))**2);         #Primary Impedance reffered to secondary\n",
+      "\n",
+      "# Calculations\n",
+      "#Distribution Tranformer Parameters\n",
+      "Rts = 1;\n",
+      "Zts = 1.9;\n",
+      "Xts = math.sqrt((Zts**2)-(Rts**2));\n",
+      "Ztspu = complex(Rts,Xts);\n",
+      "\n",
+      "Zts = Ztspu*((V/1000)**2)*10/Sts;         #Distribution Transformer Reacmath.tance\n",
+      "\n",
+      "Z1SL = (60./1000)*SSC;         #Impedance for 60 feet\n",
+      "\n",
+      "Zeq1 = PZ2+Zts+Z1SL;         #Total Impedance to the fault in secondary\n",
+      "\n",
+      "IfLL = V/abs(Zeq1);         #Fault Current At the secondary fault point F\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Impedance of the substation in ohms',\n",
+      "print (Ztp)\n",
+      "print 'b The Positive And Zero Sequence Impedances in  ohms',\n",
+      "print (Z1),\n",
+      "print (Z0)\n",
+      "print 'c The Line to Ground impedance in the primary system in ohms',\n",
+      "print (ZG)\n",
+      "print 'd The Total Impedance through the primary in ohms',\n",
+      "print (Zeq)\n",
+      "print 'e The Total Primary Impedance referred to the secondary in ohms',\n",
+      "print (PZ2)\n",
+      "print 'f The Distribution transformer impedance in ohms',\n",
+      "print (Zts)\n",
+      "print 'g the Impedance of the secondary cable in ohms',\n",
+      "print (Z1SL)\n",
+      "print 'h The Total Impedance to the fault in ohms',\n",
+      "print (Zeq1)\n",
+      "print 'i) The Single Phase line to line fault for the 120/240 V three-wire service in amperes is %g A'%(IfLL)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Impedance of the substation in ohms (0.207334533333+1.43645578359j)\n",
+        "b The Positive And Zero Sequence Impedances in  ohms (0.087+0.1812j) (0.1653+0.4878j)\n",
+        "c The Line to Ground impedance in the primary system in ohms (0.1131+0.2834j)\n",
+        "d The Total Impedance through the primary in ohms (0.320434533333+1.71985578359j)\n",
+        "e The Total Primary Impedance referred to the secondary in ohms (0.00035603837037+0.00191095087065j)\n",
+        "f The Distribution transformer impedance in ohms (0.00576+0.00930556478673j)\n",
+        "g the Impedance of the secondary cable in ohms (0.02796+0.001758j)\n",
+        "h The Total Impedance to the fault in ohms (0.0340760383704+0.0129745156574j)\n",
+        "i) The Single Phase line to line fault for the 120/240 V three-wire service in amperes is 6582.1 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch11_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch11_1.ipynb
new file mode 100755
index 00000000..3fce23ef
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch11_1.ipynb
@@ -0,0 +1,606 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:12638b4b6feef6ce42affa16d2293ad388bb21932c76952667badfebd6be692b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Distribution System Reliability"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1  Page No : 598"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "Rsys = 0.99         #Minimum Acceptable System Reliabilty\n",
+      "n = 15.;         #Number of identical Components\n",
+      "\n",
+      "# Calculations\n",
+      "q = (1-Rsys)/n;         #Probability of component failure\n",
+      "Ri = 1-q;         #Approximate value of the component reliability\n",
+      "\n",
+      "# Results\n",
+      "print 'The Approximate Value of The component reliability is %g'%(Ri)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Approximate Value of The component reliability is 0.999333\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2  Page No : 606"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "L = 4.;         #Total Length of the cable\n",
+      "Lov = 3.;        #Length of Overhead Cable\n",
+      "Lu = L-Lov;         #Length of Underground Cable\n",
+      "Nct = 2.;         #Number of circuit terminations\n",
+      "T = 10.;         #No of years for which the record is shown\n",
+      "\n",
+      "Fov = 2.;         # Faults Per Mile of the Over Head Cable\n",
+      "Fu = 1.;         #Faults Per Mile of The Underground cable\n",
+      "\n",
+      "Ct = 0.3/100        # Cable Termination Fault Rate\n",
+      "\n",
+      "#Repair Time\n",
+      "Tov = 3.;         #Over Head\n",
+      "Tu = 28.;         #Underground\n",
+      "Tct = 3.;         #Cable Termination\n",
+      "\n",
+      "# Calculations\n",
+      "lamdaFDR =  (Lov*Fov/T)+(Lu*Fu/T)+(2*Ct);         #Total Annual Fault Rate\n",
+      "rFDR = ((Tov*Lov*Fov/T)+(Tu*Lu*Fu/T)+(2*Ct*Tct))/lamdaFDR;         #Annual Fault Restoration Time\n",
+      "mFDR = 8760-rFDR;         #Annual Mean Time of Failure\n",
+      "UFDR = rFDR*100/(rFDR+mFDR);         #Unavailability of Feeder\n",
+      "AFDR = 100-UFDR;         #Availability of Feeder\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Total Annual Fault Rate is %g faults per year'%(lamdaFDR)\n",
+      "print 'b) The Annual Fault Restoration Time is %g hours per fault per year'%(rFDR)\n",
+      "print 'c) Unavailability of the feeder is %g percent'%(UFDR)\n",
+      "print 'd) Availability of the feeder is %g percent'%(AFDR)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Total Annual Fault Rate is 0.706 faults per year\n",
+        "b) The Annual Fault Restoration Time is 6.54108 hours per fault per year\n",
+        "c) Unavailability of the feeder is 0.0746698 percent\n",
+        "d) Availability of the feeder is 99.9253 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3  Page No : 608"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Annual average Fault rates\n",
+      "Fm = 0.08;\n",
+      "Fl = 0.2;\n",
+      "\n",
+      "\n",
+      "#Average Repair Times\n",
+      "Rm = 3.5;         #Main\n",
+      "Rl = 1.5;         #Lateral\n",
+      "Rs = 0.75;        #Manual Sections\n",
+      "\n",
+      "# Dismath.tances of the Lateral Feeders of A,B, and C respectively\n",
+      "Lla = 2.;\n",
+      "Llb = 1.5;\n",
+      "Llc = 1.5;\n",
+      "\n",
+      "# Dismath.tances of the Main Feeders of A,B, and C respectively\n",
+      "Lma = 1.;\n",
+      "Lmb = 1.;\n",
+      "Lmc = 1.;\n",
+      "\n",
+      "# Calculations\n",
+      "TFm = (Lma*Fm)+(Lmc*Fm)+(Lmb*Fm);         #Annual Fault of the Main Sections\n",
+      "\n",
+      "def SusInt(y): \n",
+      "    return TFm+(Fl*y)\n",
+      "\n",
+      "#Sustained Interruption Rates for A,B and C\n",
+      "IrA = SusInt(Lla);\n",
+      "IrB = SusInt(Llb);\n",
+      "IrC = SusInt(Llc);\n",
+      "\n",
+      "#Annual Repair time for A,B and C\n",
+      "rA = ((Lma*Fm*Rm)+(Lmb*Fm*Rs)+(Lmc*Fm*Rs)+(Lla*Fl*Rl))/IrA;\n",
+      "rB = ((Lma*Fm*Rm)+(Lmb*Fm*Rm)+(Lmc*Fm*Rs)+(Llb*Fl*Rl))/IrB;\n",
+      "rC = ((Lma*Fm*Rm)+(Lmb*Fm*Rm)+(Lmc*Fm*Rm)+(Llc*Fl*Rl))/IrC;\n",
+      "\n",
+      "# Results\n",
+      "print 'i The Annual Sustained Interruption Rates for:'\n",
+      "print 'Customer A : %g faults per year'%(IrA)\n",
+      "print 'Customer B : %g faults per year'%(IrB)\n",
+      "print 'Customer C : %g faults per year'%(IrC)\n",
+      "print 'ii The Average Annual Repair Time Restoration Time for:'\n",
+      "print 'Customer A : %g hours per fault per year'%(rA)\n",
+      "print 'Customer A : %g hours per fault per year'%(rB)\n",
+      "print 'Customer A : %g hours per fault per year'%(rC)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i The Annual Sustained Interruption Rates for:\n",
+        "Customer A : 0.64 faults per year\n",
+        "Customer B : 0.54 faults per year\n",
+        "Customer C : 0.54 faults per year\n",
+        "ii The Average Annual Repair Time Restoration Time for:\n",
+        "Customer A : 1.5625 hours per fault per year\n",
+        "Customer A : 1.98148 hours per fault per year\n",
+        "Customer A : 2.38889 hours per fault per year\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4  Page No : 612"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Ri = 0.85;\n",
+      "\n",
+      "def relp(y,z):\n",
+      "    return 1-((1-(Ri**y))**z)        #Equal Parallel Combination\n",
+      "\n",
+      "def rels(y,z):\n",
+      "    return (1-((1-Ri)**y))**z         #Equal Series Combination\n",
+      "\n",
+      "# Calculations\n",
+      "#Case 1: 4 elements in series\n",
+      "\n",
+      "Req1 =  rels(1,4);\n",
+      "\n",
+      "#Case 2: Two Comination of 4 elements in series, parallel to each other\n",
+      "\n",
+      "Req2 = relp(4,2);\n",
+      "\n",
+      "#Case 3 : ((two elements in series)        #(two elements in series))in series with ((two elements in series)        #(two elements in series))\n",
+      "\n",
+      "#Two Segments\n",
+      "R1 = relp(2,2);\n",
+      "R2 = relp(2,2);\n",
+      "Req3 = R1*R2;\n",
+      "\n",
+      "#Case 4 : (two elements in parallel)in series with ((three elements in series)        #(three elements in series))\n",
+      "\n",
+      "#Two Segments\n",
+      "R1 = relp(1,2);\n",
+      "R2 = relp(3,2);\n",
+      "Req4 = R1*R2;\n",
+      "\n",
+      "#Case 5, 4 groups of (2 elements in parallel) connected in series to each other\n",
+      "Req5 = rels(2,4);\n",
+      "\n",
+      "# Results\n",
+      "print 'The Equivalent System reliability for:'\n",
+      "print 'a) Configuration A : %g'%(Req1)\n",
+      "print 'b) Configuration B : %g'%(Req2)\n",
+      "print 'c) Configuration C : %g'%(Req3)\n",
+      "print 'd) Configuration D : %g'%(Req4)\n",
+      "print 'e) Configuration E : %g'%(Req5)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Equivalent System reliability for:\n",
+        "a) Configuration A : 0.522006\n",
+        "b) Configuration B : 0.771522\n",
+        "c) Configuration C : 0.851917\n",
+        "d) Configuration D : 0.831951\n",
+        "e) Configuration E : 0.912992\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5  Page No : 614"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Design the system to meet the given Equivalent System Reliability\n",
+      "#Page 614\n",
+      "\n",
+      "# Variables\n",
+      "#Individual System Reliabilities\n",
+      "Ra = 0.8;\n",
+      "Rb = 0.95;\n",
+      "Rc = 0.99;\n",
+      "Rd = 0.90;\n",
+      "Re = 0.65;\n",
+      "\n",
+      "# Calculations\n",
+      "#When All Are Connected in Series\n",
+      "\n",
+      "Req = Ra*Rb*Rc*Rd*Re;         #Equivalent System Reliability\n",
+      "\n",
+      "Rr = 0.8;         #Required\n",
+      "\n",
+      "Rae = Rr/(Rb*Rc*Rd);\n",
+      "\n",
+      "#Since Connecting the elements in parallel will increase their reliability\n",
+      "def rel(Ri,y,):\n",
+      "    return  (1-((1-Ri)**y))         #Equal Only Parallel Combination\n",
+      "\n",
+      "#Since Connecting the elements in parallel will increase their reliability\n",
+      "#Conditions to Find The Number of Elements to be used\n",
+      "for i in range(1,11):\n",
+      "    L = i;         #Number of Time Element A is used\n",
+      "    R1 = rel(Ra,i);\n",
+      "    X = R1-Rae;\n",
+      "    if(abs(X)+X == 0):\n",
+      "        continue;\n",
+      "    else:\n",
+      "        break;\n",
+      "\n",
+      "for i in range(1,11):\n",
+      "    M = i;        #Number of Time Element E is used\n",
+      "    R2 = rel(Re,i);\n",
+      "    X = R2-Rae;\n",
+      "    if(abs(X)+X == 0):\n",
+      "        continue;\n",
+      "    else:\n",
+      "        break;\n",
+      "\n",
+      "print 'a) The Equivalent system Reliability is %g'%(Req)\n",
+      "print 'b) One Each of B,C and D all connected in series are connected in serieswith the series combination of XComination of\\\n",
+      " %g elements of A, \\nAll Connected in Parallel)and YComination of %g elements of E, \\nAll Connected in Parallel) to\\\n",
+      "  achieve %g Equivalent System Realibility'%(L,M,Rr)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Equivalent system Reliability is 0.440154\n",
+        "b) One Each of B,C and D all connected in series are connected in serieswith the series combination of XComination of 2 elements of A, \n",
+        "All Connected in Parallel)and YComination of 3 elements of E, \n",
+        "All Connected in Parallel) to  achieve 0.8 Equivalent System Realibility\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6  Page No : 614"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Find The Probability on the reliability of transformers\n",
+      "#Page 614\n",
+      "\n",
+      "# Variables\n",
+      "#Reliabilities of The Three Transformers\n",
+      "Pa = 0.9;\n",
+      "Pb = 0.95;\n",
+      "Pc = 0.99;\n",
+      "\n",
+      "#Faliures of Three Transformers\n",
+      "Qa = 1-Pa;\n",
+      "Qb = 1-Pb;\n",
+      "Qc = 1-Pc;\n",
+      "\n",
+      "# Calculations\n",
+      "#Probability of NO Transformer Failing\n",
+      "Pnf = Pa*Pb*Pc;\n",
+      "\n",
+      "PfA = Qa*Pb*Pc        #Probability of Transformer A Failing\n",
+      "PfB = Pa*Qb*Pc        #Probability of Transformer B Failing\n",
+      "PfC = Pa*Pb*Qc        #Probability of Transformer C Failing\n",
+      "\n",
+      "PfAB = Qa*Qb*Pc        #Probability of Transformer A and B Failing\n",
+      "PfBC = Pa*Qb*Qc        #Probability of Transformer B and C Failing\n",
+      "PfCA = Qa*Pb*Qc        #Probability of Transformer C and A Failing\n",
+      "\n",
+      "Pf = Qa*Qb*Qc;         #Probability of All Transformers failing\n",
+      "\n",
+      "# Results\n",
+      "print 'a) Probability of No Transformer Failing is %g'%(Pnf)\n",
+      "print 'b'\n",
+      "print 'Probability of Transformer A Failing is %g'%(PfA)\n",
+      "print 'Probability of Transformer B Failing is %g'%(PfB)\n",
+      "print 'Probability of Transformer C Failing is %g'%(PfC)\n",
+      "print 'c'\n",
+      "print 'Probability of Transformers A and B Failing is %g'%(PfAB)\n",
+      "print 'Probability of Transformers B and C Failing is %g'%(PfBC)\n",
+      "print 'Probability of Transformers C and A Failing is %g'%(PfCA)\n",
+      "print 'd) Probability of All Three Transformers Failing is %g'%(Pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) Probability of No Transformer Failing is 0.84645\n",
+        "b\n",
+        "Probability of Transformer A Failing is 0.09405\n",
+        "Probability of Transformer B Failing is 0.04455\n",
+        "Probability of Transformer C Failing is 0.00855\n",
+        "c\n",
+        "Probability of Transformers A and B Failing is 0.00495\n",
+        "Probability of Transformers B and C Failing is 0.00045\n",
+        "Probability of Transformers C and A Failing is 0.00095\n",
+        "d) Probability of All Three Transformers Failing is 5e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7  Page No : 619"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Determine Probabilities Using Markovian Principle\n",
+      "#Page 619\n",
+      "\n",
+      "# Variables\n",
+      "#Conditional Probabilites Present Future\n",
+      "Pdd = 2./100;         #Down Down\n",
+      "Pud = 5./100;         #Up Down\n",
+      "Pdu = 1-Pdd;         #Down up\n",
+      "Puu = 1-Pud;         #Up Up\n",
+      "\n",
+      "# Calculations\n",
+      "P = [[Pdd,Pdu],[Pud,Puu]];         #Transition Matrix\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Conditional Probabilites for'\n",
+      "print 'Transformers Down in Present and Down in Future is %g'%(Pdd)\n",
+      "print 'Transformers Down in Present and Up in Future is %g'%(Pdd)\n",
+      "print 'Transformers Up in Present and Down in Future is %g'%(Pdd)\n",
+      "print 'Transformers Up in Present and Up in Future is %g'%(Pdd)\n",
+      "print 'b The Transition Matrix is',\n",
+      "print (P)\n",
+      "print 'c The Transition Diagram can be viewed with the result file attached to this code'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Conditional Probabilites for\n",
+        "Transformers Down in Present and Down in Future is 0.02\n",
+        "Transformers Down in Present and Up in Future is 0.02\n",
+        "Transformers Up in Present and Down in Future is 0.02\n",
+        "Transformers Up in Present and Up in Future is 0.02\n",
+        "b The Transition Matrix is [[0.02, 0.98], [0.05, 0.95]]\n",
+        "c The Transition Diagram can be viewed with the result file attached to this code\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8  Page No : 620"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Determine the Conditional Outage Probabilites\n",
+      "#Page 620\n",
+      "\n",
+      "# Calculations\n",
+      "#Conditional Outage Probabilites From The Table Given\n",
+      "P11 = 40./100;\n",
+      "P12 = 30./100;\n",
+      "P13 = 30./100;\n",
+      "P21 = 20./100;\n",
+      "P22 = 50./100;\n",
+      "P23 = 30./100;\n",
+      "P31 = 25./100;\n",
+      "P32 = 25./100;\n",
+      "P33 = 50./100;\n",
+      "\n",
+      "#Transition Matrix\n",
+      "P = [[P11,P12,P13],[P21,P22,P23],[P31,P32,P33]];\n",
+      "\n",
+      "print \"a The Conditional Outage Probabilites for:\"\n",
+      "print \"Presently Outaged Feeder is 1, Next Outaged Feeder is 1 is %g\"%(P11)\n",
+      "print \"Presently Outaged Feeder is 1, Next Outaged Feeder is 2 is %g\"%(P12)\n",
+      "print \"Presently Outaged Feeder is 1, Next Outaged Feeder is 3 is %g\"%(P13)\n",
+      "print \"Presently Outaged Feeder is 2, Next Outaged Feeder is 1 is %g\"%(P21)\n",
+      "print \"Presently Outaged Feeder is 2, Next Outaged Feeder is 2 is %g\"%(P22)\n",
+      "print \"Presently Outaged Feeder is 2, Next Outaged Feeder is 3 is %g\"%(P23)\n",
+      "print \"Presently Outaged Feeder is 3, Next Outaged Feeder is 1 is %g\"%(P31)\n",
+      "print \"Presently Outaged Feeder is 3, Next Outaged Feeder is 2 is %g\"%(P32)\n",
+      "print \"Presently Outaged Feeder is 3, Next Outaged Feeder is 3 is %g\"%(P33)\n",
+      "print \"b Transition Matrix is\"\n",
+      "print (P)\n",
+      "print \"c The Transition figure is print layed in the result file attached to this code\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Conditional Outage Probabilites for:\n",
+        "Presently Outaged Feeder is 1, Next Outaged Feeder is 1 is 0.4\n",
+        "Presently Outaged Feeder is 1, Next Outaged Feeder is 2 is 0.3\n",
+        "Presently Outaged Feeder is 1, Next Outaged Feeder is 3 is 0.3\n",
+        "Presently Outaged Feeder is 2, Next Outaged Feeder is 1 is 0.2\n",
+        "Presently Outaged Feeder is 2, Next Outaged Feeder is 2 is 0.5\n",
+        "Presently Outaged Feeder is 2, Next Outaged Feeder is 3 is 0.3\n",
+        "Presently Outaged Feeder is 3, Next Outaged Feeder is 1 is 0.25\n",
+        "Presently Outaged Feeder is 3, Next Outaged Feeder is 2 is 0.25\n",
+        "Presently Outaged Feeder is 3, Next Outaged Feeder is 3 is 0.5\n",
+        "b Transition Matrix is\n",
+        "[[0.4, 0.3, 0.3], [0.2, 0.5, 0.3], [0.25, 0.25, 0.5]]\n",
+        "c The Transition figure is print layed in the result file attached to this code\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9  Page No : 624"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array\n",
+      "\n",
+      "# Variables\n",
+      "P = array([[0.6,0.4],[0.3,0.7]]);         #One Step Transition Matrix\n",
+      "\n",
+      "Po = array([0.8,0.2]);         #Initial State Probability Vector\n",
+      "\n",
+      "# Calculations\n",
+      "#Funtion to determine the Vector of State Probability\n",
+      "def VSP(y): \n",
+      "    return (Po*(P**y))\n",
+      "\n",
+      "P1 = VSP(1);         #Vector of State Probability at Time t1\n",
+      "P4 = VSP(4);         #Vector of State Probability at Time t4\n",
+      "P8 = VSP(8);         #Vector of State Probability at Time t8\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Vector of State Probability at time t1 is',\n",
+      "print (P1)\n",
+      "print 'a The Vector of State Probability at time t4 is',\n",
+      "print (P4)\n",
+      "print 'a The Vector of State Probability at time t8 is',\n",
+      "print (P8)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Vector of State Probability at time t1 is [[ 0.48  0.08]\n",
+        " [ 0.24  0.14]]\n",
+        "a The Vector of State Probability at time t4 is [[ 0.10368  0.00512]\n",
+        " [ 0.00648  0.04802]]\n",
+        "a The Vector of State Probability at time t8 is [[  1.34369280e-02   1.31072000e-04]\n",
+        " [  5.24880000e-05   1.15296020e-02]]\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch2_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch2_1.ipynb
new file mode 100755
index 00000000..f2865431
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch2_1.ipynb
@@ -0,0 +1,861 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:45c2bf36a13c4358e12d368504a2e6b1e7cd12f738f3e5c55995bd4950ad15ca"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 :\n",
+      "Load Characteristics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1  Page No : 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "from numpy import linspace,array\n",
+      "from matplotlib.pyplot import bar,suptitle,xlabel,ylabel\n",
+      "\n",
+      "# Variables\n",
+      "t =  linspace(0,24,25)\n",
+      "SL = array([100,100,100,100,100,100,100,100,0,0,0,0,0,0,0,0,0,0,100,100,100,100,100,100,100]);\n",
+      "R = array([200,200,200,200,200,200,200,300,400,500,500,500,500,500,500,500,500,600,700,800,1000,1000,800,600,300]);\n",
+      "C = array([200,200,200,200,200,200,200,200,300,500,1000,1000,1000,1000,1200,1200,1200,1200,800,400,400,400,200,200,200]);\n",
+      "\n",
+      "# Calculations\n",
+      "Tl = SL+R+C;\n",
+      "\n",
+      "# Results\n",
+      "#To print lay the Load bar curve diagram\n",
+      "bar(t,Tl,color='red')#,0.5,'red')\n",
+      "suptitle('Example 2.1')\n",
+      "xlabel(\"Time in hrs\")\n",
+      "ylabel(\"Load in kW\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 7,
+       "text": [
+        "<matplotlib.text.Text at 0x1098963d0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x107575d10>"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2  Page No : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Fls = 0.15;\n",
+      "Ppl = 80*(10**3);         #Power Loss at peak load.\n",
+      "\n",
+      "# Calculations\n",
+      "Avgpl = Fls*Ppl;         #Average Power Loss\n",
+      "TAELCu = Avgpl*8760;         #Total annual energy loss\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The average annual power loss  =  %g kW'%(Avgpl/1000)\n",
+      "print ' b) The total annual energy loss due to the copper losses of the feeder circuits  =  %g kWh'%(TAELCu/1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The average annual power loss  =  12 kW\n",
+        " b) The total annual energy loss due to the copper losses of the feeder circuits  =  105120 kWh\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3  Page No : 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "TCDi = [9,9,9,9,9,9];         #Load for each house all in kilowatt\n",
+      "DFi = 0.65;         #Demand factor\n",
+      "Fd = 1.1;         #Diversity factor\n",
+      "\n",
+      "# Calculations\n",
+      "Dg = sum(TCDi)*DFi/Fd;\n",
+      "\n",
+      "# Results\n",
+      "print 'The diversified demand of the group on the distribution transformer is %g kW'%(Dg)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diversified demand of the group on the distribution transformer is 31.9091 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4  Page No : 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "SP = 3000*(10**3);         #System peak in kVA per phase\n",
+      "Cl = 0.5/100;         #Percentage of copper loss\n",
+      "\n",
+      "# Calculations\n",
+      "I2R =  Cl*SP;         #Copper loss of the feeder per phase\n",
+      "I2R3 = 3*I2R;         #Copper losses of the feeder per 3 phase\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The copper loss of the feeder per phase  =  %g kW'%(I2R/1000)\n",
+      "print ' b) The total coper losses of the feeder per three phase  =  %g kW'%(I2R3/1000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The copper loss of the feeder per phase  =  15 kW\n",
+        " b) The total coper losses of the feeder per three phase  =  45 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5  Page No : 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Pi = 2000.*(10**3);         #Peak for industrial load\n",
+      "Pr = 2000.*(10**3);         #Peak for residential load\n",
+      "Dg = 3000.*(10**3);         #System peak load as specified in the diagram\n",
+      "P = [Pi,Pr];         #System peaks for various loads \n",
+      "\n",
+      "# Calculations\n",
+      "Fd =  sum(P)/Dg;         #Diversity factor\n",
+      "LD =  sum(P)-Dg;         #Load diversity factor\n",
+      "Fc = 1/Fd;         # Coincidence factor\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The diversity factor of the load is %g'%(Fd)\n",
+      "print ' b) The load diversity of the load is %g kW'%(LD/1000)\n",
+      "print ' c) The coincidence factor of the load is %g'%(Fc)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The diversity factor of the load is 1.33333\n",
+        " b) The load diversity of the load is 1000 kW\n",
+        " c) The coincidence factor of the load is 0.75\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6  Page No : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import array,transpose\n",
+      "\n",
+      "# Variables\n",
+      "#Refer diagram of the first example of this chapter\n",
+      "Ps = 100.;         #Peak load for street lighting in kW\n",
+      "Pr = 1000.;         #Peak load for Residential in kW\n",
+      "Pc = 1200.;         #Peak Commercial load in kW\n",
+      "P = array([Ps,Pr,Pc])         #Peaks of various loads\n",
+      "\n",
+      "Ls5 = 0.;         #Street lighting load at 5.00 PM in kW\n",
+      "Lr5 = 600.;         #Residential load at 5.00 PM in kW\n",
+      "Lc5 = 1200.;         #Commercial Load at 5.00 PM in kW\n",
+      "\n",
+      "# Calculations\n",
+      "Cstreet = Ls5/Ps;\n",
+      "Cresidential = Lr5/Pr;\n",
+      "Ccommercial = Lc5/Pc;\n",
+      "C = array([Cstreet,Cresidential,Ccommercial]);         #Class distribution for various factors\n",
+      "\n",
+      "Fd = (sum(P))/(sum(P* transpose(C)));\n",
+      "Dg = (sum(P* transpose(C)));\n",
+      "Fc = 1/Fd;\n",
+      "\n",
+      "print 'a The class distribution factors factor of:'\n",
+      "print ' i) Street lighting  =  %g \\\n",
+      "\\nii) Residential  =  %g \\\n",
+      "\\niii) Commercial  = %g'%(C[0],C[1],C[2])\n",
+      "print ' b) The diversity factor for the primary feeder  =  %g'%(Fd)\n",
+      "print ' c) The diversified maximum demand of the load group  =  %g kW'%(Dg) \n",
+      "print ' d) The coincidence factor of the load group  =  %g'%(Fc)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The class distribution factors factor of:\n",
+        " i) Street lighting  =  0 \n",
+        "ii) Residential  =  0.6 \n",
+        "iii) Commercial  = 1\n",
+        " b) The diversity factor for the primary feeder  =  1.27778\n",
+        " c) The diversified maximum demand of the load group  =  1800 kW\n",
+        " d) The coincidence factor of the load group  =  0.782609\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7  Page No : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print 'Assuming a monthly load curve as shown in the figure attached\\\n",
+      " to this code'\n",
+      "\n",
+      "# Variables\n",
+      "TAE = 10.**7;         # Total annual energy in kW\n",
+      "APL = 3500.;         #Annual peak load in kW\n",
+      "\n",
+      "# Calculations\n",
+      "Pav =  TAE/8760;         #Annual average power demand\n",
+      "Fld =  Pav/APL;         #Annual load factor\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The annual power demand is %g kW'%(Pav)\n",
+      "print 'b) The annual load factor is %g'%(Fld)\n",
+      "print 'The unsold energy, as shown in the figure is a measure of capacity and investment math.cost. Ideally it should be kept at a minimum'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Assuming a monthly load curve as shown in the figure attached to this code\n",
+        "a) The annual power demand is 1141.55 kW\n",
+        "b) The annual load factor is 0.326158\n",
+        "The unsold energy, as shown in the figure is a measure of capacity and investment math.cost. Ideally it should be kept at a minimum\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8  Page No : 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "print 'Assuming a monthly load curve as shown in the figure attached to this code'\n",
+      "\n",
+      "# Variables\n",
+      "Nl = 100.;         #100% percent load to be supplied\n",
+      "TAE = 10.**7;         # Total annual energy in kW\n",
+      "APL = 3500.;         #Annual peak load in kW\n",
+      "\n",
+      "# Calculations\n",
+      "Pav =  TAE/8760;         #Annual average power demand\n",
+      "Fld =  (Pav+Nl)/(APL+Nl);         #Annual load factor\n",
+      "Cr = 3;        #Capacity math.cost\n",
+      "Er = 0.03;        #Energy math.cost\n",
+      "ACC = Nl*12*Cr;         #Additional capacity math.cost per kWh\n",
+      "AEC = Nl*8760*Er;         #Additional energy math.cost per kWh\n",
+      "TAC = ACC+AEC;         #Total annual math.cost\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The new annual load factor on the substation is %g'%(Fld)\n",
+      "print 'b) The total annual additional costs to NL&NP to serve this load is $%g'%(TAC)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Assuming a monthly load curve as shown in the figure attached to this code\n",
+        "a) The new annual load factor on the substation is 0.344876\n",
+        "b) The total annual additional costs to NL&NP to serve this load is $29880\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9  Page No : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "TAE = 5.61*(10**6);         #Total annual energy in kW\n",
+      "APL = 2000.;         #Annual peak load in kW\n",
+      "Lc = 0.03;         #Cost of energy per kWh in dollars\n",
+      "Plp = 100.;         #Power at peak load in kW\n",
+      "\n",
+      "# Calculations\n",
+      "Fld = TAE/(APL*8760); \n",
+      "Fls =  (0.3*Fld)+(0.7*(Fld**2));\n",
+      "AvgEL = Fls*Plp;         #Average energy loss\n",
+      "AEL = AvgEL*8760;         #Annual energy loss\n",
+      "Tlc = AEL*Lc;         #Cost of total annual copper loss\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The annual loss factor is %g'%(Fls)\n",
+      "print ' b) The annual copper loss energy is %g kWh and the cost of total annual copper loss is $%g'%(AEL,Tlc)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The annual loss factor is 0.167834\n",
+        " b) The annual copper loss energy is 147022 kWh and the cost of total annual copper loss is $4410.67\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10  Page No : 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array,arctan,transpose,arccos,radians\n",
+      "from sympy import Symbol,solve\n",
+      "\n",
+      "Fd = 1.15;\n",
+      "Pi = [1800,2000,2200];         #Demands of various feeders in kW (Real Power)\n",
+      "PF = [0.95,0.85,0.90];         #Power factor of the respective feeders\n",
+      "Dg = sum(Pi)/Fd;\n",
+      "P = Dg;\n",
+      "theta = arccos(PF);\n",
+      "\n",
+      "Q = sum(Pi*(radians(arctan(theta))))/Fd;\n",
+      "S = math.sqrt((P**2)+(Q**2));\n",
+      "LD = sum(Pi)-Dg;\n",
+      "\n",
+      "        #Transformer sizes\n",
+      "Tp = array([2500,3750,5000,7500]);\n",
+      "Ts = array([3125,4687,6250,9375]); \n",
+      "\n",
+      "Ol = 1.25;         #Maximum overload condition\n",
+      "Eol = Ts*Ol;         #Overload voltages of the transformer\n",
+      "Ed = abs(Eol-S);         # Difference between the overload values of the transformers and the P value of the system\n",
+      "\n",
+      "A = sorted(Ed);         # To sort the differences and choose the best match\n",
+      "\n",
+      "T = [Tp[1],Ts[1]];         #Suitable transformer\n",
+      "\n",
+      "g = Symbol('g');\n",
+      "X = (1+g)-pow(2,1./10);         #To find out the fans on rating\n",
+      "R = solve(X)[0];\n",
+      "g = R*100;\n",
+      "\n",
+      "n = Symbol('n');\n",
+      "Sn = 9375.;         # Rating of the to be installed transformer\n",
+      "        # Equation (1+g)**n * S  =  Sn\n",
+      "        # a = (1+g)\n",
+      "        # b = Sn/S\n",
+      "\n",
+      "a = 1+R;\n",
+      "b = Sn/S;\n",
+      "n = math.log(b)/math.log(a);\n",
+      "\n",
+      "print 'a) The 30 mins annual maximum deman on the substation transformer are %g kW and %g kVA respectively'%(P,S)\n",
+      "print ' b) The load diversity is %g kW'%(LD)\n",
+      "print ' c) Suitable transformer size for 25 percent short time over loads is %g/%g kVA'%(T[0],T[1])\n",
+      "print ' d) Fans on rating is %g percent and it will loaded for %g more year if a 7599/9375 kVA transformer is installed'%(g,math.ceil(n))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The 30 mins annual maximum deman on the substation transformer are 5217.39 kW and 5217.53 kVA respectively\n",
+        " b) The load diversity is 782.609 kW\n",
+        " c) Suitable transformer size for 25 percent short time over loads is 3750/4687 kVA\n",
+        " d) Fans on rating is 7.17735 percent and it will loaded for 9 more year if a 7599/9375 kVA transformer is installed\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11  Page No : 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import array\n",
+      "\n",
+      "print 'NOTE'\n",
+      "print 'The figure 1 attached along with this code is the Maximum diversified 30- min demand characteristics\\\n",
+      "of various residential loads;  A  =  Clothes dryer; D  =  range; E  =  lighting and miscellaneous appliances; \\\n",
+      "J  =  refrigerator Only the loads required for this problem have been mentioned '\n",
+      "\n",
+      "Ndt = 50.;         #Number of distribution transformers\n",
+      "Nr = 900.;         #Number of residences\n",
+      "\n",
+      "        #When the loads are six.\n",
+      "PavMax6 = array([1.6,0.8,0.066,0.61]);         #Average Maximum diversified demands (in kW) per house for dryer, range, refrigerator, for lighting and misc aapliances respectively according to the figure 1 attached with code. \n",
+      "\n",
+      "Mddt =  sum(6*PavMax6);         #30 min maximum diversified demand on the distribution transformer\n",
+      "\n",
+      "        #When the loads are 900.\n",
+      "PavMax900 = array([1.2,0.53,0.044,0.52]);         #        #Average Maximum diversified demands (in kW) per house for dryer, range, refreigerato, for lighting and misc aapliances respectively according to the figure 1 attached with code.\n",
+      "\n",
+      "Mdf = sum(Nr*PavMax900);         #30 min maximum diversified demand on the feeder\n",
+      "\n",
+      "        #From the figure 2 attached to this code\n",
+      "Hdd4 = array([0.38,0.24,0.9,0.32]);         #Hourly variation factor at time 4 PM for dryer, range, refrigerator, lighting and misc appliances\n",
+      "Hdd5 = array([0.30,0.80,0.9,0.70]);         #Hourly variation factor at time 5 PM for dryer, range, refrigerator, lighting and misc appliances\n",
+      "Hdd6 = array([0.22,1.0,0.9,0.92]);         #Hourly variation factor at time 6 PM for dryer, range, refrigerator, lighting and misc appliances\n",
+      "\n",
+      "Thdd4 = (6*PavMax6)*(Hdd4.T);         #Gives the total hourly diversified demand in kW at time 4 PM\n",
+      "Thdd5 = (6*PavMax6)*Hdd5;         #Gives the total hourly diversified demand in kW at time 5 PM\n",
+      "Thdd6 = (6*PavMax6)*Hdd6;         #Gives the total hourly diversified demand in kW at time 6 PM\n",
+      "\n",
+      "print ' a) The 30 min maximum diversified demand on the distribution transformer  =  %g kW'%(Mddt)\n",
+      "print ' b) The 30 min maximum diversified demand on the distribution transformer  =  %g kW'%(Mdf)\n",
+      "print ' c The total hourly diversified demands at:'\n",
+      "print ' i) 4.00 PM  =  %g kW'%(sum(Thdd4))\n",
+      "print ' ii) 5.00 PM  =  %g kW'%(sum(Thdd5))\n",
+      "print ' iii) 6.00 PM  =  %g kW'%(sum(Thdd6))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NOTE\n",
+        "The figure 1 attached along with this code is the Maximum diversified 30- min demand characteristicsof various residential loads;  A  =  Clothes dryer; D  =  range; E  =  lighting and miscellaneous appliances; J  =  refrigerator Only the loads required for this problem have been mentioned \n",
+        " a) The 30 min maximum diversified demand on the distribution transformer  =  18.456 kW\n",
+        " b) The 30 min maximum diversified demand on the distribution transformer  =  2064.6 kW\n",
+        " c The total hourly diversified demands at:\n",
+        " i) 4.00 PM  =  6.3276 kW\n",
+        " ii) 5.00 PM  =  9.6384 kW\n",
+        " iii) 6.00 PM  =  10.6356 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12  Page No : 72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy import array,arctan,radians,degrees,ceil\n",
+      "\n",
+      "T = 730;         #Average monthly time in hrs\n",
+      "Pla = 22;         #Peak Load for consumer A in kW\n",
+      "Plb = 39;         #Peak load for consumer B in kW\n",
+      "MEC = array([0.025,0.02,0.015]);         #Monthly Energy charges in cents/kWh according to the units consumed\n",
+      "Uc = array([1000,3000,3000]);         #Units consumption according to the Energy charges\n",
+      "MDC = 2;         #Monthly demand charge in dollars/kW\n",
+      "\n",
+      "Pa = 7000.;         #Units served to Consumer A in kWh\n",
+      "Pb = 7000.;         #Units served to Consumer B in kWh\n",
+      "\n",
+      "#Power factors\n",
+      "Pfa = 0.9;         # Lag\n",
+      "Pfb = 0.76;         #Lag\n",
+      "\n",
+      "#Monthly Load factors\n",
+      "Flda = Pa/(Pla*T);\n",
+      "Fldb = Pb/(Plb*T);\n",
+      "\n",
+      "#Continous kilovoltamperes for each distribution transformer\n",
+      "Sa = Pla/Pfa;\n",
+      "Sb = Plb/Pfb;\n",
+      "\n",
+      "#Ratings of the distribution transformers needed\n",
+      "Ta = round(Sa/5)*5;\n",
+      "Tb = round(Sb/5)*5;\n",
+      "\n",
+      "#Billing Charges\n",
+      "#For Consumer A\n",
+      "Mbda = Pla*(0.85/Pfa);         # Monthly billing demand\n",
+      "Mdca = Mbda*MDC;         #Monthly demand charge\n",
+      "#Since the units served are 7000 it can be split according to the rates  as 1000, 3000, 3000 excess units.\n",
+      "Uca = Uc;         #Units consumption by A\n",
+      "Meca = MEC*Uca.T;         #Monthly energy charge\n",
+      "Tmba = Meca+Mdca;         #Total monthly bill\n",
+      "\n",
+      "#For Consumer B\n",
+      "Mbdb = Plb*(0.85/Pfb);         # Monthly billing demand\n",
+      "Mdcb = Mbdb*MDC;         #Monthly demand charge\n",
+      "#Since the units served are 7000 it can be split according to the rates  as 1000, 3000, 3000 excess units.\n",
+      "Ucb = Uc;         #Units consumption by B\n",
+      "Mecb = MEC*Ucb.T;         #Monthly energy charge\n",
+      "Tmbb = Mecb+Mdcb;         #Total monthly bill\n",
+      "\n",
+      "#To find the capacitor size\n",
+      "Q1 = Pb*math.radians(arctan(math.acos(Pfb)));         #For original power factor\n",
+      "Q2 = Pb*math.radians(arctan(math.acos(0.85)));         #For new power factor\n",
+      "\n",
+      "dQ = (Q1-Q2)/T;         #Capacitor size\n",
+      "\n",
+      "#For new power factor\n",
+      "#For Consumer B\n",
+      "Mbdbn = Plb*(1);         # Monthly billing demand\n",
+      "Mdcbn = Mbdbn*MDC;         #Monthly demand charge\n",
+      "#Since the units served are 7000 it can be split according to the rates  as 1000, 3000, 3000 excess units.\n",
+      "Ucbn = Uc;         #Units consumption by B\n",
+      "Mecbn = MEC*Ucbn.T;         #Monthly energy charge\n",
+      "Tmbbn = Mecbn+Mdcbn;         #Total monthly bill\n",
+      "\n",
+      "Saving = abs(Tmbbn-Tmbb);         #Saving due to capacitor installation\n",
+      "Ci = 30;         # Cost of capacitor in dollar per kVAr\n",
+      "Cc = Ci*dQ;         #The math.cost of the installed capacitor\n",
+      "PP = Cc/Saving;         #Payback Period\n",
+      "PPr = 90/Saving;         #Realistic Payback period\n",
+      "\n",
+      "print 'a Monthly load factor for :'\n",
+      "print ' i) Consumer A  =  %g'%(Flda)\n",
+      "print ' ii) Consumer B  = %g'%(Fldb)\n",
+      "print ' b Rating of the each of the distribution transformer:'\n",
+      "print ' i) A  =  %g kVA'%(Ta)\n",
+      "print ' ii) B  =  %g kVA'%(Tb)\n",
+      "print ' c Monthly bil for:'\n",
+      "print ' i) Consumer A  =  $%g'%(sum(Tmba)) \n",
+      "print ' ii) Consumer B  =  $%g'%(sum(Tmbb))\n",
+      "print ' d) The capacitor size required is %g kVAr'%(dQ)\n",
+      "print ' e Payback period:'\n",
+      "print ' i) Calculated : %g months'%(sum(ceil(PP)))\n",
+      "print ' ii) Realistic as capacitor size available is 3 kVAr : %g months'%(sum(ceil(PPr)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a Monthly load factor for :\n",
+        " i) Consumer A  =  0.435866\n",
+        " ii) Consumer B  = 0.245873\n",
+        " b Rating of the each of the distribution transformer:\n",
+        " i) A  =  25 kVA\n",
+        " ii) B  =  50 kVA\n",
+        " c Monthly bil for:\n",
+        " i) Consumer A  =  $254.667\n",
+        " ii) Consumer B  =  $391.711\n",
+        " d) The capacitor size required is 0.018276 kVAr\n",
+        " e Payback period:\n",
+        " i) Calculated : 3 months\n",
+        " ii) Realistic as capacitor size available is 3 kVAr : 30 months\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13  Page No : 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "Kh = 7.2;         #Meter consmath.tant\n",
+      "Kr1 = 32;         #Revolutions of the disk in the first reading\n",
+      "Kr2 = 27;         #Revolutions of the disk in the second reading\n",
+      "T1 = 59;         #Time interval for revolutions of disks for the first reading\n",
+      "T2 = 40;         #Time interval for revolutions of disks for the second reading\n",
+      "\n",
+      "# Calculations\n",
+      "# Self contained watthour meter; D  =  (3.6*Kr*Kh)/T\n",
+      "\n",
+      "def  Id1(a,b):\n",
+      "    return ((3.6*Kh*a)/b)         #Function to calculate the instaneous demand\n",
+      "\n",
+      "D1 = Id1(Kr1,T1);\n",
+      "D2 = Id1(Kr2,T2);\n",
+      "Dav = (D1+D2)/2;\n",
+      "\n",
+      "# Results\n",
+      "print 'The instantenous demands are %g kW and %g kW for reading 1 and 2 and the average demand is %g kW'%(D1,D2,Dav)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The instantenous demands are 14.0583 kW and 17.496 kW for reading 1 and 2 and the average demand is 15.7772 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14  Page No : 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#For a transformer type watthour meter; D  =  (3.6*Kr*Kh*CTR*PTR)/T\n",
+      "CTR = 200.;\n",
+      "PTR = 1.;\n",
+      "Kh = 1.8;\n",
+      "Kr1 = 32.;         #Revolutions of the disk in the first reading\n",
+      "Kr2 = 27.;         #Revolutions of the disk in the second reading\n",
+      "T1 = 59.;         #Time interval for revolutions of disks for the first reading\n",
+      "T2 = 40.;         #Time interval for revolutions of disks for the second reading\n",
+      "\n",
+      "# Calculations\n",
+      "def Id1(a,b):\n",
+      "    return ((3.6*Kh*a*CTR*PTR)/b)         #Function to calculate the instaneous demand\n",
+      "\n",
+      "D1 = Id1(Kr1,T1);\n",
+      "D2 = Id1(Kr2,T2);\n",
+      "Dav = (D1+D2)/2;\n",
+      "\n",
+      "# Results\n",
+      "print 'The instantenous demands are %g kW and %g kW for reading 1 and 2 and the average demand is %g kW'%(D1,D2,Dav)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The instantenous demands are 702.915 kW and 874.8 kW for reading 1 and 2 and the average demand is 788.858 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15  Page No : 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Kh = 1.2;\n",
+      "CTR = 80.;\n",
+      "PTR = 20.;\n",
+      "#Revolutions of the disk in a watthour meter for 1 and 2 readings respectively\n",
+      "Kr1 = 20.;\n",
+      "Kr2 = 30.;\n",
+      "#Revolutions of the disk in a VArhour meter for 1 and 2 readings respectively\n",
+      "Kr3 = 10.;\n",
+      "Kr4 = 20.\n",
+      "#Time interval for revoltion of disks in watthour meter for 1 and 2 readings respectively\n",
+      "T1 = 50.;\n",
+      "T2 = 60.;\n",
+      "#Time interval for revoltion of disks in VArhour meter for 1 and 2 readings respectively\n",
+      "T3 = 50.;\n",
+      "T4 = 60.;\n",
+      "\n",
+      "def Id1(a,b):\n",
+      "    return ((3.6*Kh*a*CTR*PTR)/b)         #Function to calculate the instaneous demand\n",
+      "\n",
+      "#instanmath.taneous kilowatt demands for readings 1 and 2\n",
+      "D1 = Id1(Kr1,T1);\n",
+      "D2 = Id1(Kr2,T2);\n",
+      "\n",
+      "#instanmath.taneous kilovar deamnds for readings 1 and 2\n",
+      "D3 = Id1(Kr3,T3);\n",
+      "D4 = Id1(Kr4,T4);\n",
+      "\n",
+      "Davp = (D1+D2)/2;         #Average kilowatt demand\n",
+      "Davq = (D3+D4)/2;         #Average kilovar demand\n",
+      "\n",
+      "Dav = math.sqrt((Davp**2)+(Davq**2));         #Average kilovoltampere demand\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The instanmath.taneous kilowatt hour demands for readings 1 and 2 are %g kW and %g kW respectively'%(D1,D2)\n",
+      "print ' b) The average kilowatt demand is %g kW'%(Davp)\n",
+      "print ' c) The instanmath.taneous kilovar hour demands for readings 1 and 2 are %g kVAr and %g kVAr respectively'%(D3,D4)\n",
+      "print ' d) The average kilovar demand is %g kVAr'%(Davq)\n",
+      "print ' e) The average kilovoltampere demand is %g kVA'%(Dav)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The instanmath.taneous kilowatt hour demands for readings 1 and 2 are 2764.8 kW and 3456 kW respectively\n",
+        " b) The average kilowatt demand is 3110.4 kW\n",
+        " c) The instanmath.taneous kilovar hour demands for readings 1 and 2 are 1382.4 kVAr and 2304 kVAr respectively\n",
+        " d) The average kilovar demand is 1843.2 kVAr\n",
+        " e) The average kilovoltampere demand is 3615.52 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch3_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch3_1.ipynb
new file mode 100755
index 00000000..60be3cd4
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch3_1.ipynb
@@ -0,0 +1,877 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f499e025f89e71e922f5b5ff841986e94328c0797b70d556107927513cde3781"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Application of Distribution Transformers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1  Page No : 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "S = 25.*(10**3);         #Rating of the transformer in VA\n",
+      "#Values in per unit\n",
+      "Rt = 0.014;         #Resismath.tance of Transformer\n",
+      "Xt = 0.012;         #Reacmath.tance of transformer\n",
+      "Vh = 7200;         #High Voltage End in V\n",
+      "Vx = 120;         # Low Voltage End in V\n",
+      "Rb = (Vh**2)/S;         #Base Value of Resismath.tance\n",
+      "#Accroding to Lloyd's Formula\n",
+      "\n",
+      "# Calculations\n",
+      "Zhx12 = (1.5*Rt)+(1j*1.2*Xt);         #Impedance referred to HV side when the winding x2x3 is shorted\n",
+      "\n",
+      "n = Vh/Vx;         #Turns Ratio\n",
+      "\n",
+      "Zhx13 = Rt+(1j*Xt);         #Use of Entire low voltage winding\n",
+      "\n",
+      "#Impedances of the required terms in pu\n",
+      "A = (2*Zhx13)-Zhx12;\n",
+      "B = ((2*Zhx12)-(2*Zhx13))/(n**2);\n",
+      "C = B;\n",
+      "\n",
+      "#Angle of Impedances\n",
+      "ta = math.degrees(math.atan(A.imag/A.real));\n",
+      "tb = math.degrees(math.atan(B.imag/B.real));\n",
+      "\n",
+      "# Results\n",
+      "print 'The Circuit impedances on the high voltage side is %g/_%g ohm'%(abs(A*Rb),ta)\n",
+      "print 'Each of the Circuit impedances on the low voltage side is %g/_%g ohm'%(abs(B*Rb),tb)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Circuit impedances on the high voltage side is 24.6366/_53.9017 ohm\n",
+        "Each of the Circuit impedances on the low voltage side is 0.0085248/_18.9246 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2  Page No : 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "#Impedances from the previous example\n",
+      "Zh = 24.6437*exp(1j*53.9*math.pi/180);\n",
+      "Zl = 8.525*(10**-3)*exp(1j*18.9*math.pi/180);\n",
+      "#Voltages\n",
+      "Vh = 7200.;         #High End\n",
+      "Vx = 120.;         # Low End\n",
+      "S = 25.*1000;         #Transformer Rating in VA\n",
+      "N = Vh/Vx;         #Turns Ratio\n",
+      "\n",
+      "# Calculations\n",
+      "#R of service drop is zero         #Line to Neutral Currents\n",
+      "IfLVn = Vx/(Zl+((1/(N**2))*Zh));         #Secondary Fault Current\n",
+      "IfHVn = IfLVn/N;         #Primary Fault Current\n",
+      "\n",
+      "#R of service drop is zero         #Line to Line Currents\n",
+      "Nl = Vh/(2*Vx);         #New Truns Ratio\n",
+      "IfLVl = 2*Vx/((2*Zl)+((1/(Nl**2))*Zh));         #Secondary Fault Current\n",
+      "IfHVl = IfLVl/Nl;         #Primary Fault Current\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop\\\n",
+      " is zero are %g A and %g A respectively'%(abs(IfHVn),abs(IfLVn))\n",
+      "print 'b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are\\\n",
+      " %g A and %g A respectively'%(abs(IfHVl),abs(IfLVl))\n",
+      "print 'c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is %g A'%(abs(IfLVn))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop is zero are 136.353 A and 8181.2 A respectively\n",
+        "b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are 188.283 A and 5648.5 A respectively\n",
+        "c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is 8181.2 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3  Page No : 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy import solve,Symbol\n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "Vx = 120.;         #Low End Voltage\n",
+      "#When Service drop is Zero\n",
+      "IfLVn = 8181.7*exp(-1*1j*34.3*math.pi/180);         #Line to Neutral Vault Current\n",
+      "IfLVl = 5649*exp(-1*1j*40.6*math.pi/180);         #Line to Line Fault Current\n",
+      "\n",
+      "Ral4 = 2.58;         #        #4 AWG Aluminium Conductor Resismath.tance per mile\n",
+      "Ralinf = 1.03;         #        #1/0 AWG Aluminium Conductor Resismath.tance per mile\n",
+      "\n",
+      "# Calculations\n",
+      "#Impedances when Service drop is zero, suffix l denotes line to line\n",
+      "#Suffix n denotes line to neutral\n",
+      "Zl0 = (2*Vx)/IfLVl;\n",
+      "Zn0 = (Vx)/IfLVn;\n",
+      "\n",
+      "#When there is R service drop\n",
+      "#Magnitudes of Line to Line and Line to Earth fault currents are equal\n",
+      "\n",
+      "R = Symbol('R');         #Variable Value\n",
+      "#Effective Impedances\n",
+      "Zl = Zl0+(2*R);\n",
+      "Zn = Zn0+(2*R);\n",
+      "#Fault Currents\n",
+      "Ifl = 2*Vx/Zl;\n",
+      "Ifn = Vx/Zn;\n",
+      "#print Ifl[1]\n",
+      "#Magnitudes of Currents\n",
+      "MIfl = abs(240.)/abs(Ifl.subs(R,3));\n",
+      "MIfn = abs(Ifn.subs(R,2))/abs(Ifn.subs(R,3));\n",
+      "DI = MIfl-MIfn;\n",
+      "X = - 1.5781966 + 240*R   #DI.subs(R,2);         #Polynomial Equation to find 'R'\n",
+      "R = solve(X)[0];         #Numerical Value\n",
+      "\n",
+      "#The Magnitude of R found is Wrong in the Textbook\n",
+      "\n",
+      "#Length of service drop cable\n",
+      "SDL4 = R/Ral4;\n",
+      "SDLinf = R/Ralinf;\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Value of Service drop in the Cable is %g ohm'%(R)\n",
+      "print 'b The Length of service drop cable for:'\n",
+      "print 'i)         #4 AWG Conductor is %g miles'%(SDL4) \n",
+      "print 'ii)         #1/0 AWG Conductor is %g miles'%(SDLinf) \n",
+      "\n",
+      "#Length is printed in Miles\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Value of Service drop in the Cable is 0.00657582 ohm\n",
+        "b The Length of service drop cable for:\n",
+        "i)         #4 AWG Conductor is 0.00254877 miles\n",
+        "ii)         #1/0 AWG Conductor is 0.00638429 miles\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4  Page No : 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#Transformer Ratings in kVA\n",
+      "Sr1 = 250.; \n",
+      "Sr2 = 500.;\n",
+      "\n",
+      "#percentage impedances\n",
+      "Zr1 = 2.4;\n",
+      "Zr2 = 3.1;\n",
+      "\n",
+      "# Calculations\n",
+      "#Ratio of Maximum Loads\n",
+      "R = Sr1*Zr2/(Sr2*Zr1);\n",
+      "\n",
+      "#If 500 kVA is chosen as the full load transformer, Transformer 1 becomes overloaded\n",
+      "SL1 = Sr1;         #To Avoid OverLoading of transformer 1\n",
+      "\n",
+      "SL2 = SL1/R;         #Maximum Load on transformer 2\n",
+      "\n",
+      "Tl = SL1+SL2;         #Total Load without overloading\n",
+      "\n",
+      "# Results\n",
+      "print 'The Maximum Load Carried without overloading any of the transformer is %g kVA'%(Tl)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Maximum Load Carried without overloading any of the transformer is 637.097 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5  Page No : 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Considering Van as reference voltage\n",
+      "SL3phi = 200*(10**3);         #Load to be powered\n",
+      "pf3 = 0.8;         #Power Factor of three phase load\n",
+      "t3 = math.acos(pf3);         #Power FActor Angle for three phase load\n",
+      "pf1 = 0.9;         #Power Factor of math.single phase load\n",
+      "t1 = math.acos(pf1);         #Power Factor angle of math.single phase load\n",
+      "SL1 = 80.*(10**3);         #Single Phase Light Load\n",
+      "Vll = 240.;         #Secondary Voltage\n",
+      "#Rating of Single Phase Transformers between individual lines\n",
+      "Sbc = 100.*(10**3);\n",
+      "Sab = 75.*(10**3);\n",
+      "Sca = Sab;\n",
+      "#Angles of Three phase voltages \n",
+      "ta = 0.;\n",
+      "tb = -120.;\n",
+      "tc = 120.;\n",
+      "#Angles of three  line currents\n",
+      "tai = ta-t3;\n",
+      "tbi = tb-t3;\n",
+      "tci = tc-t3;\n",
+      "\n",
+      "# Calculations\n",
+      "I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current\n",
+      "#3 Phase Line Currents\n",
+      "Ia3 = I*exp(1j*math.pi*tai/180);\n",
+      "Ib3 = I*exp(1j*math.pi*tbi/180);\n",
+      "Ic3 = I*exp(1j*math.pi*tci/180);\n",
+      "\n",
+      "MIbc = SL1/Vll;         #Magnitude Single Phase Current\n",
+      "\n",
+      "tbc = -90;         #Lagging Van         #Angle of Vbc\n",
+      "tbci = tbc-t1;         #Angle of Current Ibc\n",
+      "Ibc = MIbc*exp(1j*math.pi*tbci/180);\n",
+      "\n",
+      "#Load Currents\n",
+      "Ia = Ia3;\n",
+      "Ta = math.degrees(math.atan(Ia.imag/Ia.real));\n",
+      "Ib = Ib3+Ibc;\n",
+      "Tb = math.degrees(math.atan(Ib.imag/Ib.real));\n",
+      "Ic = Ic3-Ibc;         #Current is wrong in the textbook\n",
+      "Tc = math.degrees(math.atan(Ic.imag/Ic.real));\n",
+      "\n",
+      "#Current Flowing in the secondary winding of the transformers 1,2 and 3\n",
+      "Iac = ((Ic/Sbc)-(Ia/Sab))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
+      "T1 = math.degrees(math.atan(Iac.imag/Iac.real));         #Angle of the above current\n",
+      "Iba = ((Ia/Sca)-(Ib/Sbc))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
+      "T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current\n",
+      "Icb = ((Ib/Sab)-(Ic/Sca))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
+      "T3 = math.degrees(math.atan(Icb.imag/Icb.real));         #Angle of the above current\n",
+      "\n",
+      "#Kilovoltampere Load on each transformer\n",
+      "SLab = Vll*abs(Iba)/1000;\n",
+      "SLbc = Vll*abs(Icb)/1000;\n",
+      "SLca = Vll*abs(Iac)/1000;\n",
+      "\n",
+      "Vlls = Vll;         #Secondary Voltage\n",
+      "Vllp = 7620;         #Primary Voltage\n",
+      "n = Vllp/Vlls;         #Turns Ratio\n",
+      "\n",
+      "#Primary Currents of the transformer\n",
+      "IAC = Iac/n;\n",
+      "IBA = Iba/n;\n",
+      "ICB = Icb/n;\n",
+      "\n",
+      "#Primary Current in each each phase wire\n",
+      "IA = IAC-IBA;\n",
+      "TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current\n",
+      "IB = IBA-ICB;\n",
+      "TB = math.degrees(math.atan(IB.imag/IB.real));        #Angle of the above current\n",
+      "IC = ICB-IAC;\n",
+      "TC = math.degrees(math.atan(IC.imag/IC.real));        #Angle of the above current\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Line Currents flowing in secondary phase wire :'\n",
+      "print 'A phase is %g/_%g A'%(abs(Ia),Ta)\n",
+      "print 'B phase is %g/_%g A'%(abs(Ib),Tb)\n",
+      "print 'C phase is %g/_%g A'%(abs(Ic),Tc)\n",
+      "print 'b The Current flowing in secondary winding of each transformer:'\n",
+      "print 'AC is %g/_%g A'%(abs(Iac),T1)\n",
+      "print 'AB is %g/_%g A'%(abs(Iba),T2)\n",
+      "print 'BC is %g/_%g A'%(abs(Icb),T3)\n",
+      "print 'c The Load on Each Transformer is:'\n",
+      "print '1 : %g kVA'%(SLca)\n",
+      "print '2 : %g kVA'%(SLab)\n",
+      "print '3 : %g kVA'%(SLbc)\n",
+      "print 'd The Current flowing in primary winding of each transformer:'\n",
+      "print 'AC is %g/_%g A'%(abs(IAC),T1)\n",
+      "print 'AB is %g/_%g A'%(abs(IBA),T2)\n",
+      "print 'BC is %g/_%g A'%(abs(ICB),T3)\n",
+      "print 'e The Line Currents flowing in primary phase wire :'\n",
+      "print 'A phase is %g/_%g A'%(abs(IA),TA)\n",
+      "print 'B phase is %g/_%g A'%(abs(IB),TB)\n",
+      "print 'C phase is %g/_%g A'%(abs(IC),TC)\n",
+      "\n",
+      "#Ic is calculation is wrong, the author has added Ibc instead of subtracting, so if you change - into + in line 45, you get the answer as in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Line Currents flowing in secondary phase wire :\n",
+        "A phase is 481.125/_-0.643501 A\n",
+        "B phase is 787.293/_71.6504 A\n",
+        "C phase is 787.977/_-72.7823 A\n",
+        "b The Current flowing in secondary winding of each transformer:\n",
+        "AC is 316/_-40.9814 A\n",
+        "AB is 315.535/_39.7662 A\n",
+        "BC is 545.454/_89.442 A\n",
+        "c The Load on Each Transformer is:\n",
+        "1 : 75.84 kVA\n",
+        "2 : 75.7283 kVA\n",
+        "3 : 130.909 kVA\n",
+        "d The Current flowing in primary winding of each transformer:\n",
+        "AC is 9.95276/_-40.9814 A\n",
+        "AB is 9.9381/_39.7662 A\n",
+        "BC is 17.1796/_89.442 A\n",
+        "e The Line Currents flowing in primary phase wire :\n",
+        "A phase is 15.1536/_-0.643501 A\n",
+        "B phase is 24.7966/_71.6504 A\n",
+        "C phase is 24.8182/_-72.7823 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6  Page No : 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Considering Van as reference voltage\n",
+      "SL3phi = 100.*(10**3);         #Load to be powered\n",
+      "pf3 = 0.8;         #Power Factor of three phase load\n",
+      "t3 = math.acos(pf3);         #Power FActor Angle for three phase load\n",
+      "pf1 = 0.9;         #Power Factor of math.single phase load\n",
+      "t1 = math.acos(pf1);         #Power Factor angle of math.single phase load\n",
+      "SL1 = 50.*(10**3);         #Single Phase Light Load\n",
+      "Vll = 240.;         #Secondary Voltage\n",
+      "#Angles of Three phase voltages \n",
+      "ta = 0.;\n",
+      "tb = -120.;\n",
+      "tc = 120.;\n",
+      "#Angles of three  line currents\n",
+      "tai = ta-t3;\n",
+      "tbi = tb-t3;\n",
+      "tci = tc-t3;\n",
+      "\n",
+      "# Calculations\n",
+      "I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current\n",
+      "#3 Phase Line Currents\n",
+      "Ia3 = I*exp(1j*math.pi*tai/180);\n",
+      "Ib3 = I*exp(1j*math.pi*tbi/180);\n",
+      "Ic3 = I*exp(1j*math.pi*tci/180);\n",
+      "\n",
+      "MI1 = SL1/Vll;         #Magnitude Single Phase Current\n",
+      "\n",
+      "t1v = 30;         #Leading Van         #Angle of Vbc\n",
+      "t1i = t1v-t1;         #Angle of Current Ibc\n",
+      "I1 = MI1*exp(1j*math.pi*t1i/180);\n",
+      "\n",
+      "#Load Currents\n",
+      "Ia = Ia3+I1;\n",
+      "Ta = math.degrees(math.atan(Ia.imag/Ia.real));\n",
+      "Ib = Ib3-I1;\n",
+      "Tb = -180+(math.degrees(math.atan(Ib.imag/Ib.real)));\n",
+      "Ic = Ic3;         #Current is wrong in the textbook\n",
+      "Tc = math.degrees(math.atan(Ic.imag/Ic.real));\n",
+      "\n",
+      "#Current flowing in the secondary of the transformer\n",
+      "Iba = Ia;\n",
+      "T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current\n",
+      "Icb = Ic;\n",
+      "T3 = 180+(math.degrees(math.atan(Icb.imag/Icb.real)));         #Angle of the above current\n",
+      "\n",
+      "#Load on Each Transformer\n",
+      "SLba = Vll*abs(Iba)/1000;\n",
+      "SLcb = Vll*abs(Icb)/1000;\n",
+      "\n",
+      "Vlls = Vll;         #Secondary Voltage\n",
+      "Vllp = 7620;         #Primary Voltage\n",
+      "n = Vllp/Vlls;         #Turns Ratio\n",
+      "\n",
+      "#Primary Currents of the transformer\n",
+      "IA = Iba/n;\n",
+      "TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current\n",
+      "IB = Icb/n;\n",
+      "TB = T3;        #Angle of the above current\n",
+      "IN = IA+IB;         #Neutral Current\n",
+      "TN = math.degrees(math.atan(IN.imag/IN.real));        #Angle of the above current\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Line Currents flowing in secondary phase wire :'\n",
+      "print 'A phase is %g/_%g A'%(abs(Ia),Ta)\n",
+      "print 'B phase is %g/_%g A'%(abs(Ib),Tb)\n",
+      "print 'C phase is %g/_%g A'%(abs(Ic),Tc)\n",
+      "print 'b The Current flowing in secondary winding of each transformer:'\n",
+      "print 'AB is %g/_%g A'%(abs(Iba),T2)\n",
+      "print 'BC is %g/_%g A'%(abs(Icb),T3)\n",
+      "print 'c The Load on Each Transformer is:'\n",
+      "print '1 : %g kVA'%(SLba)\n",
+      "print '2 : %g kVA'%(SLcb)\n",
+      "print 'd The Line Currents flowing in primary phase wire :'\n",
+      "print 'AB is %g/_%g A'%(abs(IA),TA)\n",
+      "print 'CB is %g/_%g A'%(abs(IB),TB)\n",
+      "print 'The Neutral Current is %g/_%g'%(abs(IN),TN)\n",
+      "\n",
+      "#Note the mistake in the Textbook for the calulation for Neutral Current\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Line Currents flowing in secondary phase wire :\n",
+        "A phase is 433.486/_13.3432 A\n",
+        "B phase is 433.874/_-134.453 A\n",
+        "C phase is 240.563/_-60.6435 A\n",
+        "b The Current flowing in secondary winding of each transformer:\n",
+        "AB is 433.486/_13.3432 A\n",
+        "BC is 240.563/_119.356 A\n",
+        "c The Load on Each Transformer is:\n",
+        "1 : 104.037 kVA\n",
+        "2 : 57.735 kVA\n",
+        "d The Line Currents flowing in primary phase wire :\n",
+        "AB is 13.6531/_13.3432 A\n",
+        "CB is 7.57678/_119.356 A\n",
+        "The Neutral Current is 13.6653/_45.5475\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8  Page No : 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "Vll = 480.;         #Line to Line Voltage\n",
+      "Vln = 277.;         #Line to neutral Voltage\n",
+      "\n",
+      "# Calculations\n",
+      "#From the Phasor Diagram from the result file\n",
+      "Vab = Vll*exp(1j*0);         #Vab is taken as reference\n",
+      "Vabh = 50*Vab/100;\n",
+      "VAB = 4160.;\n",
+      "VABh = 50*VAB/100; \n",
+      "VH1H2o = math.sqrt((VAB**2)-(VABh**2));\n",
+      "VH1H2t = (VABh);\n",
+      "Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "VH2H3t = (VABh);\n",
+      "Vx1x2t = Vabh;\n",
+      "Vx2x3t = Vabh;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Phasor diagram is shown in the result file attached to the code'\n",
+      "print 'b) Vab is %g/_%g V'%(abs(Vab),Vab.imag/Vab.real)\n",
+      "print 'c The Magnitudes of the following rated winding voltages'\n",
+      "print 'i) The Voltage VH1H2 on transformer 1 : %g V'%(VH1H2o.real)\n",
+      "print 'ii) The Voltage Vx1x2 on transformer 1 : %g V'%(Vx1x2o.real)\n",
+      "print 'iii) The Voltage Vx2x3 on transformer 1 : %g V'%(Vx2x3o.real)\n",
+      "print 'iv) The Voltage VH1H2 on transformer 2 : %g V'%(VH1H2t.real)\n",
+      "print 'v) The Voltage VH2H3 on transformer 2 : %g V'%(VH2H3t.real)\n",
+      "print 'vi) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx1x2t.real)\n",
+      "print 'vii) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx2x3t.real)\n",
+      "print 'd i NO ii NO iii YES'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Phasor diagram is shown in the result file attached to the code\n",
+        "b) Vab is 480/_0 V\n",
+        "c The Magnitudes of the following rated winding voltages\n",
+        "i) The Voltage VH1H2 on transformer 1 : 3602.67 V\n",
+        "ii) The Voltage Vx1x2 on transformer 1 : 138.564 V\n",
+        "iii) The Voltage Vx2x3 on transformer 1 : 277.128 V\n",
+        "iv) The Voltage VH1H2 on transformer 2 : 2080 V\n",
+        "v) The Voltage VH2H3 on transformer 2 : 2080 V\n",
+        "vi) The Voltage Vx1x2 on transformer 2 : 240 V\n",
+        "vii) The Voltage Vx1x2 on transformer 2 : 240 V\n",
+        "d i NO ii NO iii YES\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9  Page No : 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "R = 2.77;         #Resismath.tance of the balanced load\n",
+      "#From Phasor Diagram in Result file\n",
+      "Vab = 480*exp(1j*0);         #Reference Voltage\n",
+      "MVn = abs(Vab)/math.sqrt(3);         #Magnitude of line to neutral voltages\n",
+      "#Angles of Three phase voltages \n",
+      "ta = -30.;\n",
+      "tb = -150.;\n",
+      "tc = 90.;\n",
+      "\n",
+      "# Calculations\n",
+      "#Angles of Winding according to the Line Currents\n",
+      "tx3x2 = 30;         #Leading\n",
+      "tx1x2 = -30;         #Lagging\n",
+      "\n",
+      "I = MVn/R;         #Magnitude of current\n",
+      "\n",
+      "#Low Voltage Current Phasors\n",
+      "Ia = I*exp(1j*math.pi*ta/180);\n",
+      "Ib = I*exp(1j*math.pi*tb/180);\n",
+      "Ic = I*exp(1j*math.pi*tc/180);\n",
+      "pfT = math.cos(math.radians(ta-ta));         #Angle of Ia is same as phase voltage         #Resismath.tance load\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Low voltage current phasors are:'\n",
+      "print 'A is %g/_%g A'%(abs(Ia),ta)\n",
+      "print 'B is %g/_%g A'%(abs(Ib),tb)\n",
+      "print 'C is %g/_%g A'%(abs(Ic),tc)\n",
+      "print 'b The Phasor Diagram is the ''b'' diagram of in the result file'\n",
+      "print 'c) The Power Factor of the Transformer is %g'%(pfT)\n",
+      "print 'd) Power Factor as seen by winding x3x2 of transformer 2 is %g leading'%(math.cos(math.radians(tx3x2)))\n",
+      "print 'e) Power Factor as seen by winding x1x2 of transformer 2 is %g lagging'%(math.cos(math.radians(tx1x2)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Low voltage current phasors are:\n",
+        "A is 100.046/_-30 A\n",
+        "B is 100.046/_-150 A\n",
+        "C is 100.046/_90 A\n",
+        "b The Phasor Diagram is the b diagram of in the result file\n",
+        "c) The Power Factor of the Transformer is 1\n",
+        "d) Power Factor as seen by winding x3x2 of transformer 2 is 0.866025 leading\n",
+        "e) Power Factor as seen by winding x1x2 of transformer 2 is 0.866025 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10  Page No : 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "#'o' and 't' represent transformer one and two respectively\n",
+      "# Variables\n",
+      "#Objective is to find the Factor which has to be multiplied to get VA rating\n",
+      "Vll = 1.;         #Assumption Made\n",
+      "#From the Phasor Diagram from the result file\n",
+      "Vab = Vll*exp(1j*0);         #Vab is taken as reference\n",
+      "Vabh = 50.*Vab/100;\n",
+      "Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "Vx1x2t = Vabh;\n",
+      "Vx2x3t = Vabh;\n",
+      "\n",
+      "#Let I be unity\n",
+      "I = 1;\n",
+      "\n",
+      "# Calculations\n",
+      "#VA Ratings of the respective windings\n",
+      "Sx1x2o = Vx1x2o*I;\n",
+      "Sx2x3o = Vx2x3o*I;\n",
+      "Sx1x2t = Vx1x2t*I;\n",
+      "Sx2x3t = Vx2x3t*I;\n",
+      "\n",
+      "#Total VA rating of transformer\n",
+      "S1 = Sx1x2o+Sx2x3o;\n",
+      "S2 = Sx1x2t+Sx2x3t;\n",
+      "\n",
+      "#Ratio of total rating to maximum rating\n",
+      "Rt = (S1+S2)/(math.sqrt(3)*Vll*I);\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx1x2o.real)\n",
+      "print 'b) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx2x3o.real)\n",
+      "print 'c) The Total Output from transformer 1 is %g*VI VA'%(S1.real)\n",
+      "print 'd) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx1x2t.real)\n",
+      "print 'e) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx2x3t.real)\n",
+      "print 'f) The Total Output from transformer 2 is %g*VI VA'%(S2.real)\n",
+      "print 'g) The Total Rating to the Maximum Continous Output is %g'%(Rt.real)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The voltampere raing of x1x2 of transformer 1 is 0.288675*VI VA\n",
+        "b) The voltampere raing of x1x2 of transformer 1 is 0.57735*VI VA\n",
+        "c) The Total Output from transformer 1 is 0.866025*VI VA\n",
+        "d) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA\n",
+        "e) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA\n",
+        "f) The Total Output from transformer 2 is 1*VI VA\n",
+        "g) The Total Rating to the Maximum Continous Output is 1.07735\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11  Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "#Per unit value\n",
+      "Zt = 0.01+(1j*0.03);         #Transformer impedance\n",
+      "\n",
+      "Vll = 240.;         #Secondary Voltage\n",
+      "\n",
+      "Sl = 90.;         #Lighting Load\n",
+      "pfl = 0.9;\n",
+      "tl = math.acos(pfl);\n",
+      "S = 25.;         #Balanced Load\n",
+      "pf = 0.8;\n",
+      "t = math.acos(pf);\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "def angle(y): \n",
+      "    return math.degrees(math.atan(y.imag/y.real))\n",
+      "\n",
+      "tab = 30;         #Phase angle of Vab\n",
+      "\n",
+      "Il = Sl*1000/Vll;         #Magnitude of Light Load\n",
+      "#Umath.sing the symmetrical - components theory\n",
+      "Ia1 = Il*exp(1j*math.pi*(tab-tl)/180);\n",
+      "Ta1 = angle(Ia1);         #Angle for the above current\n",
+      "Ib1 = -1*Ia1;\n",
+      "Ic1 = 0;         #Neutral Wire\n",
+      "#Angles of three line to line voltages\n",
+      "ta = 0;\n",
+      "tb = -120;\n",
+      "tc = 120;\n",
+      "\n",
+      "Ib = S*1000/(math.sqrt(3)*Vll);         #Magnitude of balanced load\n",
+      "\n",
+      "#Currents in Three phase load\n",
+      "Ta2 = ta-t;\n",
+      "Ia2 = Ib*exp(1j*math.pi*Ta2/180);\n",
+      "Tb2 = tb-t;\n",
+      "Ib2 = Ib*exp(1j*math.pi*Tb2/180);\n",
+      "Tc2 = tc-t;\n",
+      "Ic2 = Ib*exp(1j*math.pi*Tc2/180);\n",
+      "\n",
+      "#Currents in phase wire\n",
+      "Ia = Ia1+Ia2;\n",
+      "Ta = angle(Ia);         #Angle corresponding to the above angle\n",
+      "Ib = Ib1+Ib2;\n",
+      "Tb = angle(Ib);         #Angle corresponding to the above angle\n",
+      "Ic = Ic1+Ic2;\n",
+      "Tc = angle(Ic);         #Angle corresponding to the above angle\n",
+      "\n",
+      "#Transformer Loads\n",
+      "ST1 = Vll*abs(Ia)/1000;\n",
+      "T1 = 100;         #From the above value of Load, this transformer is chosen to meet the specific characteristic\n",
+      "ST1pu = ST1/T1;         #Per unit Load\n",
+      "ST2 = Vll*abs(Ic)/1000;\n",
+      "T2 = 15;         #From the above value of Load, this transformer is chosen to meet the specific characteristic\n",
+      "ST2pu = ST2/T2;         #Per unit Load\n",
+      "\n",
+      "#Transformer Power Factors\n",
+      "pfT1 = math.cos(math.radians(tab-Ta));\n",
+      "pfT2 = math.cos(math.radians(90-Tc));         #Vcb makes angle of 90\n",
+      "\n",
+      "Vh = 7200;         #High End Voltage\n",
+      "n = Vh/Vll;         #Turns Ratio\n",
+      "\n",
+      "# The Primary Line Currents\n",
+      "IA = Ia/n;\n",
+      "IB = -1*Ic/n;\n",
+      "IN = -1*(IA+IB);\n",
+      "\n",
+      "Ibase = T1*1000/Vll;         #Base Current\n",
+      "Iapu = Ia/Ibase;\n",
+      "Icpu = Ic/Ibase;\n",
+      "\n",
+      "#Phase Voltages\n",
+      "Vab = Vll*exp(1j*math.pi*tab/180); \n",
+      "Vbc = Vll*exp(-1*1j*math.pi*90/180);\n",
+      "#Per Unit Voltages\n",
+      "VANpu = (Vab/Vll)+(Iapu*Zt);\n",
+      "VBNpu = (Vbc/Vll)-(Icpu*Zt);\n",
+      "\n",
+      "#Actual Voltages\n",
+      "VAN = VANpu*Vh;\n",
+      "VBN = VBNpu*Vh;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Phasor Currents:'\n",
+      "print 'Ia is %g/_%g A'%(abs(Ia),Ta)\n",
+      "print 'Ib is %g/_%g A'%(abs(Ib),180+Tb)\n",
+      "print 'Ic is %g/_%g A'%(abs(Ic),Tc)\n",
+      "print 'b) Suitable ratings of the transformers are %g kVA and %g kVA'%(T1,T2)\n",
+      "print 'c) The Per Unit kVA load on each transformer is %g pu and %g pu'%(ST1pu,ST2pu)\n",
+      "print 'd) The power factor of output of each transformer is %g and %g both lagging'%(pfT1,pfT2)\n",
+      "print 'e The phasor currents at the high voltage leads:'\n",
+      "print 'IA is %g/_%g A'%(abs(IA),Ta)\n",
+      "print 'IB is %g/_%g A'%(abs(IB),180+angle(IB))\n",
+      "print 'IN is %g/_%g A'%(abs(IN),angle(IN))\n",
+      "print 'f) VAN is %g/_%g V and VBN is %g/_%g V'%(abs(VAN),angle(VAN),abs(VBN),angle(VBN))\n",
+      "\n",
+      "#Highly Accuracy of Answers; Upto 5 decimal Places\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Phasor Currents:\n",
+        "Ia is 428.052/_25.4972 A\n",
+        "Ib is 428.229/_213.552 A\n",
+        "Ic is 60.1407/_-60.6435 A\n",
+        "b) Suitable ratings of the transformers are 100 kVA and 15 kVA\n",
+        "c) The Per Unit kVA load on each transformer is 1.02732 pu and 0.96225 pu\n",
+        "d) The power factor of output of each transformer is 0.996914 and -0.871586 both lagging\n",
+        "e The phasor currents at the high voltage leads:\n",
+        "IA is 14.2684/_25.4972 A\n",
+        "IB is 2.00469/_119.356 A\n",
+        "IN is 14.5415/_17.5913 A\n",
+        "f) VAN is 7294.34/_31.6923 V and VBN is 7193.85/_-89.743 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch4_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch4_1.ipynb
new file mode 100755
index 00000000..01f911b9
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch4_1.ipynb
@@ -0,0 +1,565 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:245ef1b89a5bda9ee093c72396146e897e6533850c3acc1090aa40bb52c35f20"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Design of Subtransmission Lines and Distribution Substations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1  Page No : 201"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Conductor Pararmeters\n",
+      "r = 1.503;\n",
+      "xa = 0.609;\n",
+      "xd = 0.1366;\n",
+      "pf = 0.9;\n",
+      "Vb = 2400;\n",
+      "Vr = Vb;\n",
+      "x = xa+xd;\n",
+      "Kc = 0.01;         #From the Curve\n",
+      "\n",
+      "# Calculations\n",
+      "K = ((r*pf)+(x+math.sin(math.radians(math.acos(pf)))))*(1000./3)*100/(Vr*Vb);         # In Percent\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Value of Consmath.tant K is %g percent VDpu per kVA mile'%(K)\n",
+      "print 'b) From the precalculated per cent voltage drop Curve, It is found that the K is %g percent VDpu\\\n",
+      " per kVA mile which is same as the answer obtained in part a'%(Kc)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Value of Consmath.tant K is 0.0121885 percent VDpu per kVA mile\n",
+        "b) From the precalculated per cent voltage drop Curve, It is found that the K is 0.01 percent VDpu per kVA mile which is same as the answer obtained in part a\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2  Page No : 202"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "K = 0.01;         #Percentage Value\n",
+      "Sn = 500;         #Load in kVA\n",
+      "pf = 0.9;         #Lagging\n",
+      "s = 1;         #Length of the feeder\n",
+      "\n",
+      "# Calculations\n",
+      "VD = s*K*Sn;         #Voltage drop in percent\n",
+      "\n",
+      "# Results\n",
+      "print 'The Percent Voltage drop in the Main is %g percent'%(VD)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Percent Voltage drop in the Main is 5 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3  Page No : 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "K = 0.01;         #Percentage Value\n",
+      "Sn = 500;         #Load in kVA\n",
+      "pf = 0.9;         #Lagging\n",
+      "l = 1.;         #Total Length of the feeder\n",
+      "\n",
+      "# Calculations\n",
+      "s = l/2;         #effective Length of the feeder\n",
+      "VD = s*K*Sn;         #Voltage drop in percent\n",
+      "\n",
+      "# Results\n",
+      "print 'The Percent Voltage drop in the Main is %g percent'%(VD)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Percent Voltage drop in the Main is 2.5 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4  Page No : 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "K = 0.01;         #Percentage Value\n",
+      "Sn = 500;         #Load in kVA\n",
+      "pf = 0.9;         #Lagging\n",
+      "l = 1.;         #Total Length of the feeder\n",
+      "\n",
+      "# Calculations\n",
+      "s = l*2/3;         #effective Length of the feeder\n",
+      "VD = s*K*Sn;         #Voltage drop in percent\n",
+      "\n",
+      "# Results\n",
+      "print 'The Percent Voltage drop in the Main is %g percent'%(VD)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Percent Voltage drop in the Main is 3.33333 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5  Page No : 204"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#Voltage Drops in Percentage\n",
+      "VDlumped = 5.; \n",
+      "VDuniform = 2.5;\n",
+      "VDincreasing = 3.333;\n",
+      "\n",
+      "# Calculations\n",
+      "#Ratio of the percent voltage drops\n",
+      "Rlu = VDlumped/VDuniform;\n",
+      "Rli = VDlumped/VDincreasing;\n",
+      "Riu = VDincreasing/VDuniform;\n",
+      "\n",
+      "# Results\n",
+      "print 'a) Percent VDlumped  =  %g Percent VDuniform'%(Rlu)\n",
+      "print 'b) Percent VDlumped  =  %g Percent VDincreasing'%(Rli)\n",
+      "print 'c) Percent VDincreasing  =  %g Percent VDuniform'%(Riu)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) Percent VDlumped  =  2 Percent VDuniform\n",
+        "b) Percent VDlumped  =  1.50015 Percent VDincreasing\n",
+        "c) Percent VDincreasing  =  1.3332 Percent VDuniform\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6  Page No : 208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array,multiply,divide\n",
+      "\n",
+      "# Variables\n",
+      "D = [500,500,2000,2000,10000,10000,2000,2000];         #Load Densities in kVA/sq.miles\n",
+      "TAn = [6,6,3,3,1,1,15,15];         #Substation Area in sq.miles\n",
+      "VD = [3,6,3,6,3,6,3,6];         #Maximum Total Primary Feeder Voltage drops in percentage\n",
+      "Vll = [4.16,4.16,4.16,4.16,4.16,4.16,13.2,13.2];         #Base Feeder Voltage in kV\n",
+      "\n",
+      "TSn = multiply(D,TAn);         #Susbstation Load\n",
+      "#From the Graphs of feeders vs load desity in the textbook; The Number of feeders are found to be\n",
+      "\n",
+      "n = [4,2,5,3,5,4,6,5];         #No of feeders\n",
+      "\n",
+      "# Calculations\n",
+      "#Also from the graph, The characteristic or the feeder is determined\n",
+      "#1-5, 7 are VDL feeders\n",
+      "#6 and 8 are TL feeders\n",
+      "\n",
+      "Sn = divide(TSn,n);         #Load Per Feeder\n",
+      "#To Determine the Load Current\n",
+      "Il = Sn/(math.sqrt(3)*array(Vll)); \n",
+      "\n",
+      "# Results\n",
+      "print 'a'\n",
+      "print 'The Substation Size is'\n",
+      "print (TSn)\n",
+      "print 'The Number of Feeders from the Curve is'\n",
+      "print (n)\n",
+      "print 'Also From the Curve%( 1,2,3,4,5,7 cases are VDL but 6 and 8 case are TL'\n",
+      "print 'a'\n",
+      "print 'The Load Current for 6th Case is %g A, which is less than the ampacities of the main but\\\n",
+      " more than that of the lateral, Hence it is thermally limited but not the main feeder'%(Il[5])\n",
+      "print 'The Load Current for 8th Case is %g A, which is less than the ampacities of the main but more than that\\\n",
+      " of the lateral, Hence it is thermally limited but not the main feeder'%(Il[7])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a\n",
+        "The Substation Size is\n",
+        "[ 3000  3000  6000  6000 10000 10000 30000 30000]"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "The Number of Feeders from the Curve is\n",
+        "[4, 2, 5, 3, 5, 4, 6, 5]\n",
+        "Also From the Curve%( 1,2,3,4,5,7 cases are VDL but 6 and 8 case are TL\n",
+        "a\n",
+        "The Load Current for 6th Case is 346.965 A, which is less than the ampacities of the main but more than that of the lateral, Hence it is thermally limited but not the main feeder\n",
+        "The Load Current for 8th Case is 262.432 A, which is less than the ampacities of the main but more than that of the lateral, Hence it is thermally limited but not the main feeder\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7  Page No : 211"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "D = 1000.;         #Load Density in kVA per sq miles\n",
+      "Vll = 4.16;         #Line to Lien voltage in kV\n",
+      "#From The Tables and Curves from the Theory\n",
+      "K = 0.007;\n",
+      "#For TL\n",
+      "TLImax = 230.;         #Maximum Feeder Current\n",
+      "TLSn = math.sqrt(3)*Vll*TLImax;         #Maximum Load Per Feeder\n",
+      "TLn = 4;         #No of Feeders\n",
+      "TLTSn = TLn*TLSn;         #Substation Load\n",
+      "TLl4 = math.sqrt(TLSn/D);         #Feeder Length\n",
+      "TLS = 2*TLl4;         #Total Spacing\n",
+      "\n",
+      "TLVDn = 2*K*D*(TLl4**3)/3;         #TotalVoltageDrop in the main\n",
+      "\n",
+      "# Calculations\n",
+      "#For VDL\n",
+      "VDLVDn = 3;         #Percent Voltage Drop\n",
+      "VDLl4 = pow((3*VDLVDn/(2*K*D)),1./3);         #Feeder Length\n",
+      "VDLS = 2*VDLl4;         #Station size\n",
+      "VDLSn = D*(VDLl4**2);         #Maximum Load Per Feeder\n",
+      "VDLn = TLn;         #Number Of Feeders\n",
+      "VDLTSn = VDLn*VDLSn;         #Susbtation Load\n",
+      "VDLImax = VDLSn/(math.sqrt(3)*Vll);         #Ampere Rating of the Main\n",
+      "R = VDLImax/TLImax;         #Ampere Loading\n",
+      "\n",
+      "# Results\n",
+      "print 'a For Thermally Limited '\n",
+      "print 'i) The Substation Size  =  %g kVA'%(TLTSn)\n",
+      "print 'ii) Substation Spacing  =  %g miles'%(TLS)\n",
+      "print 'iii) Maximum Load Per Feeder  =  %g kVA'%(TLSn)\n",
+      "print 'iv) The Voltage Drop is %g percent'%(TLVDn)\n",
+      "\n",
+      "print 'b For Voltage Drop Limited '\n",
+      "print 'i) The Substation Size  =  %g kVA'%(VDLTSn)\n",
+      "print 'ii) Substation Spacing  =  %g miles'%(VDLS)\n",
+      "print 'iii) Maximum Load Per Feeder  =  %g kVA'%(VDLSn)\n",
+      "print 'iv) Ampere Loading of the Main is %g pu'%(R)\n",
+      "\n",
+      "#Note The Approximation to 750 kVA\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a For Thermally Limited \n",
+        "i) The Substation Size  =  6628.9 kVA\n",
+        "ii) Substation Spacing  =  2.57467 miles\n",
+        "iii) Maximum Load Per Feeder  =  1657.23 kVA\n",
+        "iv) The Voltage Drop is 9.95588 percent\n",
+        "b For Voltage Drop Limited \n",
+        "i) The Substation Size  =  2979.45 kVA\n",
+        "ii) Substation Spacing  =  1.72611 miles\n",
+        "iii) Maximum Load Per Feeder  =  744.863 kVA\n",
+        "iv) Ampere Loading of the Main is 0.449464 pu\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8  Page No : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "DivF = 1.2;         #Diversity Factor\n",
+      "DemF = 0.6;         #Demand Factor\n",
+      "CL = 2000.;         #Connected Load Density in kVA per sq.miles\n",
+      "\n",
+      "DD = DemF*CL/DivF;         #Diversified Demand\n",
+      "A = 4.;         #Area of the Substation\n",
+      "\n",
+      "TSn = DD*A;         #Peak Loads of A and B\n",
+      "Sm = TSn;         #Peak Loads\n",
+      "\n",
+      "#Consmath.tants for different conductors\n",
+      "Km = 0.0004;\n",
+      "Kl = 0.00095;\n",
+      "#Number of Laterals\n",
+      "Na = 16.;         #Site A \n",
+      "Nb = 32.;         #Site B\n",
+      "\n",
+      "#Length of the Main\n",
+      "La = 2.;\n",
+      "Lb = 3.;\n",
+      "#length of laterals\n",
+      "Lla = 2.;\n",
+      "Llb = 1.;\n",
+      "\n",
+      "# Calculations\n",
+      "#Length of expres Load\n",
+      "Le = 1;\n",
+      "Leffb = Le+((Lb-Le)/2);         #Effective Length of the feeder in site B\n",
+      "#Voltage drops\n",
+      "VDa = (La*Km*Sm/2)+(Lla*Kl*Sm/(Na*2));\n",
+      "VDb = (Leffb*Km*Sm)+(Llb*Kl*Sm/(Nb*2));\n",
+      "\n",
+      "# Results\n",
+      "print 'The Voltage drop in Site A is %g percent'%(VDa)\n",
+      "print 'The Voltage drop in Site B is %g percent'%(VDb)\n",
+      "VDb = (La*Km*Sm/2)+(Lla*Kl*Sm/Na);\n",
+      "print 'Comparing the results we find Site A suitable due to its less percent voltage drop',VDb\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Voltage drop in Site A is 1.8375 percent\n",
+        "The Voltage drop in Site B is 3.25938 percent\n",
+        "Comparing the results we find Site A suitable due to its less percent voltage drop 2.075\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10  Page No : 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "D = 500.;         #Load Density in kVA per sq.miles\n",
+      "Vl = 12.47;         #Line Voltage in kV\n",
+      "N = 2.;         #Feeders per substation\n",
+      "#From Table A-5 Appendix A it Current Ampacity can be found\n",
+      "\n",
+      "Imax = 340.;\n",
+      "\n",
+      "# Calculations\n",
+      "S2 = math.sqrt(3)*Vl*Imax;         #Load Per Feeder\n",
+      "\n",
+      "l2 = math.sqrt(S2/D);         #Length of the feeder\n",
+      "S = 2*l2;         #Substation Spacing\n",
+      "TS2 = S2*N;         #Total Load on substation\n",
+      "\n",
+      "# Results\n",
+      "print 'The Parts a%(b and c of thhis question cannot be coded'\n",
+      "print 'd) The substation size and spacing is %g kVA and %g miles'%(TS2,S)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Parts a%(b and c of thhis question cannot be coded\n",
+        "d) The substation size and spacing is 14687.1 kVA and 7.66475 miles\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11  Page No : 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy import Symbol,solve\n",
+      "\n",
+      "# Variables\n",
+      "Ts = 1.;         #Assumed Load on station\n",
+      "K = 1.;         #Assumption Consmath.tant\n",
+      "K2 = K;\n",
+      "K4 = K;\n",
+      "D = 1.;        #Assumption Load Density\n",
+      "#Number of feeders\n",
+      "N2 = 2.;\n",
+      "N4 = 4.;\n",
+      "S2 = Ts/N2;         #Load per feeder         #Two feeders\n",
+      "S4 = Ts/N4;         #Load per feeder         #4 feeders\n",
+      "l = Symbol('l');         #Variable Value of length\n",
+      "L2eff = 1*l/3;\n",
+      "L4eff = 2*l/3;\n",
+      "\n",
+      "# Calculations\n",
+      "def VD(y): \n",
+      "    return D*(l**2)*K*y\n",
+      "\n",
+      "VD2 = VD(L2eff);\n",
+      "VD4 = VD(L4eff);\n",
+      "RVD = VD2/VD4;\n",
+      "X = l-RVD;\n",
+      "RVD = 2.      #1./(solve(X,2)[0]);         #To find the ratio of (l2**3)/(l4**3)\n",
+      "\n",
+      "Rl = pow(RVD,1./3);         #Ratio of length of feeder for 2 feeders two by length of feeder for 4 feeders\n",
+      "\n",
+      "#A is directly proportional to l**2\n",
+      "RA = (Rl**2);\n",
+      "\n",
+      "#TSn is directly proportional to n and l**2\n",
+      "RTS = (N2/N4)*(Rl**2);\n",
+      "\n",
+      "# Results\n",
+      "print 'a) Ratio of substation spacings  =  2l2/2l4  =  %g'%(Rl)\n",
+      "print 'b) Ratio of areas covered per feeder main  =  A2/A4  =  %g'%(RA)\n",
+      "print 'c) Ratio of substation loads  =  TS2/TS4  =  %g'%(RTS)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) Ratio of substation spacings  =  2l2/2l4  =  1.25992\n",
+        "b) Ratio of areas covered per feeder main  =  A2/A4  =  1.5874\n",
+        "c) Ratio of substation loads  =  TS2/TS4  =  0.793701\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch5_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch5_1.ipynb
new file mode 100755
index 00000000..65d25e8a
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch5_1.ipynb
@@ -0,0 +1,422 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b2ff16e1aeeea980b2eb9e21bf5ebcd0fe195cb024eff9d74d8ab7ee94189858"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Design Considerations of Primary Systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1  Page No : 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp,sqrt\n",
+      "\n",
+      "# Variables\n",
+      "Z = 0.1+(0.1*1j);         #Feeder Impedance per unit\n",
+      "R = Z.real;         #Resismath.tance\n",
+      "X = Z.imag;         #Reacmath.tance\n",
+      "Vs = 1.;         #Sending End Voltage\n",
+      "Pr = 1.;         #Consmath.tant Power Load\n",
+      "pfr = 0.8;         #Power Factor at recieving end\n",
+      "tr = math.acos(pfr);         #Power FActor angle\n",
+      "\n",
+      "# Calculations\n",
+      "def angle(y): \n",
+      "    return math.degrees(math.atan(y.imag/y.real))\n",
+      "\n",
+      "K = (Vs**2)-(2*Pr*(R+(X*(math.atan(tr)))));\n",
+      "\n",
+      "Vr = math.sqrt((K/2)*(1+math.sqrt(1-((2*abs(Z)*Pr/(K*pfr))**2))));         #Recieving End Voltage\n",
+      "C = Pr*(X-(R*math.degrees(math.atan(tr))))/((Vr**2)+(Pr*(R+(X*math.degrees(math.atan(tr))))));\n",
+      "\n",
+      "del1 = math.degrees(math.atan(C));\n",
+      "\n",
+      "Ir = (Pr/(abs(Vr)*pfr))*exp(-1*math.pi*1j*tr/180)         #Recieving End Current\n",
+      "Is = Ir;         #Sending End Current\n",
+      "Tir = angle(Ir);\n",
+      "\n",
+      "Vr1 = Vs-(Z*Ir);\n",
+      "\n",
+      "# Results\n",
+      "print 'a) Vr is %g/_%g pu, del is %g degrees, Ir  =  Is  =  %g/_%g pu'%(abs(Vr.real),angle(Vr),del1,abs(Ir),Tir)\n",
+      "print 'b) Vr is %g/_%g pu, which is almost equal to the previous case.'%(Vr1.real,angle(Vr1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) Vr is 0.79784/_0 pu, del is -38.3625 degrees, Ir  =  Is  =  1.56673/_-0.643501 pu\n",
+        "b) Vr is 0.841577/_-10.4293 pu, which is almost equal to the previous case.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2  Page No : 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Sl = 518.;         #Total Load on Lateral\n",
+      "Sm = 1036.;         #Total Load on Main\n",
+      "Vll = 4.16;         #Line to Line voltage\n",
+      "\n",
+      "# Calculations\n",
+      "#Currents in the respective current\n",
+      "Ilateral = Sl/(math.sqrt(3)*Vll);\n",
+      "Imain = Sm/(math.sqrt(3)*Vll);\n",
+      "\n",
+      "C = 5280.;         #Length Consmath.tant\n",
+      "Ll = 5760./C;         #Lateral Length\n",
+      "Lm = 3300./C;         #Main Length\n",
+      "\n",
+      "#Consmath.tant for the cables\n",
+      "Kl = 0.015;\n",
+      "Km = 0.01;\n",
+      "\n",
+      "#Voltage Drop Percents for 3 phase\n",
+      "VDlateral3 = Ll*Kl*Sl/2;\n",
+      "VDmain3 = Lm*Km*Sm;\n",
+      "TVD3 = VDmain3+VDlateral3;\n",
+      "#Voltage Drop Percents for 1 phase according to Morrisoncfor laterals\n",
+      "VDlateral1 = VDlateral3*4;\n",
+      "VDmain1 = VDmain3;\n",
+      "TVD1 = VDlateral1+VDmain1;\n",
+      "\n",
+      "\n",
+      "#CASE B\n",
+      "#To meet the maximum primary voltage drop criterion of 4.00 percent\n",
+      "#Conductors with ampacities of 480A and 270A for Main and laterals\n",
+      "\n",
+      "#Consmath.tants from the table\n",
+      "Klb = 0.006;\n",
+      "Kmb = 0.003;\n",
+      "\n",
+      "#Voltage Drop Percents\n",
+      "VDlateralb = Ll*Klb*Sl/2;\n",
+      "VDmainb = Lm*Kmb*Sm;\n",
+      "TVDb = VDmainb+VDlateralb;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The percent voltage drops at :'\n",
+      "print 'i 3Phase'\n",
+      "print 'Lateral End is %g percent'%(VDlateral3)\n",
+      "print 'Main End is %g percent'%(VDmain3)\n",
+      "print 'ii 1Phase'\n",
+      "print 'Lateral End is %g percent'%(VDlateral1)\n",
+      "print 'Main End is %g percent'%(VDmain1)\n",
+      "print 'b) Conductors with Ampacities of 480A and 270A are used to find the Percent voltage drop of the \\\n",
+      "Main and Lateral as %g percent and %g percent respectively'%(VDmainb,VDlateralb)\n",
+      "print 'The Above Drops meet the required criterion of 4 percent voltage drop'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The percent voltage drops at :\n",
+        "i 3Phase\n",
+        "Lateral End is 4.23818 percent\n",
+        "Main End is 6.475 percent\n",
+        "ii 1Phase\n",
+        "Lateral End is 16.9527 percent\n",
+        "Main End is 6.475 percent\n",
+        "b) Conductors with Ampacities of 480A and 270A are used to find the Percent voltage drop of the Main and Lateral as 1.9425 percent and 1.69527 percent respectively\n",
+        "The Above Drops meet the required criterion of 4 percent voltage drop\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3  Page No : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Terms taken from Example two\n",
+      "Il = 72.; \n",
+      "Im = 144.; \n",
+      "C = 5280.;         #Length Consmath.tant\n",
+      "Ll = 5760./C;         #Lateral Length\n",
+      "Lm = 3300./C;         #Main Length\n",
+      "\n",
+      "#From Tables\n",
+      "#Lateral\n",
+      "rl = 4.13;         #Resismath.tance per mile\n",
+      "xLl = 0.258;         #Reacmath.tance per mile\n",
+      "#Main\n",
+      "rm = 1.29;         #Resismath.tance per mile\n",
+      "xLm = 0.211;        #Reacmath.tance per mile\n",
+      "pf = 0.9;         #Power Factor\n",
+      "\n",
+      "Vb = 2400.;         #Base Voltage\n",
+      "\n",
+      "# Calculations\n",
+      "#Voltage Drops\n",
+      "VDlateral = Il*((rl*pf)+(xLl*math.sin(math.radians(math.acos(pf)))))*Ll/2; \n",
+      "VDmain = Im*((rm*pf)+(xLm*math.sin(math.radians(math.acos(pf)))))*Lm;\n",
+      "\n",
+      "#Percent Voltage Drop\n",
+      "perVDlateral = VDlateral*100/Vb;\n",
+      "perVDmain = VDmain*100/Vb;\n",
+      "\n",
+      "TVD = perVDlateral+perVDmain;         #Total Percent Voltage drop\n",
+      "\n",
+      "#Case B\n",
+      "#Conductors With Ampacities of 268A and 174A for Main and Laterals\n",
+      "#From Tables\n",
+      "#Lateral\n",
+      "rlb = 1.03;         #Resismath.tance per mile\n",
+      "xLlb = 0.207;         #Reacmath.tance per mile\n",
+      "#Main\n",
+      "rmb = 0.518;         #Resismath.tance per mile\n",
+      "xLmb = 0.191;        #Reacmath.tance per mile\n",
+      "\n",
+      "Vb = 2400;         #Base Voltage\n",
+      "#Voltage Drops\n",
+      "VDlateralb = Il*((rlb*pf)+(xLlb*math.sin(math.radians(math.acos(pf)))))*Ll/2; \n",
+      "VDmainb = Im*((rmb*pf)+(xLmb*math.sin(math.radians(math.acos(pf)))))*Lm;\n",
+      "\n",
+      "#Percent Voltage Drop\n",
+      "perVDlateralb = VDlateralb*100/Vb;\n",
+      "perVDmainb = VDmainb*100/Vb;\n",
+      "\n",
+      "TVDb = perVDlateralb+perVDmainb;         #Total Percent Voltage drop\n",
+      "\n",
+      "# Results\n",
+      "print 'a The percent voltage drops at :'\n",
+      "print 'Lateral End is %g percent'%(perVDlateral)\n",
+      "print 'Main End is %g percent'%(perVDmain)\n",
+      "\n",
+      "print 'b) Conductors with Ampacities of 278A and 174A are used to find the Percent voltage drop of \\\n",
+      "the Main and Lateral as %g percent and %g percent respectively'%(perVDmainb,perVDlateralb)\n",
+      "print 'The Above Drops meet the required criterion of 4 percent voltage drop'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The percent voltage drops at :\n",
+        "Lateral End is 6.08569 percent\n",
+        "Main End is 4.35998 percent\n",
+        "b) Conductors with Ampacities of 278A and 174A are used to find the Percent voltage drop of the Main and Lateral as 1.75389 percent and 1.51958 percent respectively\n",
+        "The Above Drops meet the required criterion of 4 percent voltage drop\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4  Page No : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Sl = 518.;         #Total Load on Lateral\n",
+      "Sm = 5180.;         #Total Load on Main\n",
+      "Vll = 12.47;         #Line to Line voltage\n",
+      "\n",
+      "#Currents in the respective current\n",
+      "Ilateral = Sl/(math.sqrt(3)*Vll);\n",
+      "Imain = Sm/(math.sqrt(3)*Vll);\n",
+      "\n",
+      "C = 5280.;         #Length Consmath.tant\n",
+      "Ll = 5760./C;         #Lateral Length\n",
+      "Lm = 3300./C;         #Main Length\n",
+      "\n",
+      "#Consmath.tant for the cables\n",
+      "Km = 0.0008;\n",
+      "Kl = 0.00175;\n",
+      "\n",
+      "# Calculations\n",
+      "#Voltage Drop Percents for 3 phase\n",
+      "VDlateral = Ll*Kl*Sl/2;\n",
+      "\n",
+      "#Due to peculiarity of this new problem, one half of the main has to considered as express feeder and the other connected to a uniformly distributed load of 5180kVA\n",
+      "VDmain = Lm*Km*Sm*3/4;\n",
+      "TVD = VDmain+VDlateral;\n",
+      "\n",
+      "#Since the inductive reacmath.tance of the line is\n",
+      "Cd = 12.;         #Consmath.tant to find the dismath.tance in terms of feet\n",
+      "\n",
+      "#Diameters of the Conductors\n",
+      "Dmi = 37.;\n",
+      "Dmn = 53.;\n",
+      "\n",
+      "#Drops per mile\n",
+      "xdi = 0.1213*math.log(Dmi/Cd);\n",
+      "xdn = 0.1213*math.log(Dmn/Cd);\n",
+      "\n",
+      "Dxd = xdn-xdi;         #Difference in Drops\n",
+      "\n",
+      "# Results\n",
+      "print 'a The percent voltage drops at :'\n",
+      "print 'Lateral End is %g percent'%(VDlateral)\n",
+      "print 'Main End is %g percent'%(VDmain)\n",
+      "\n",
+      "print 'b The Above Drops meet the required criterion of 4 percent voltage drop'\n",
+      "print 'c) The Difference in Voltage drop is %g ohm/mile, which is a smaller VD valuue that it really is.'%(Dxd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The percent voltage drops at :\n",
+        "Lateral End is 0.494455 percent\n",
+        "Main End is 1.9425 percent\n",
+        "b The Above Drops meet the required criterion of 4 percent voltage drop\n",
+        "c) The Difference in Voltage drop is 0.0435921 ohm/mile, which is a smaller VD valuue that it really is.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5  Page No : 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Vb = 7200.;         #Base Voltage in V\n",
+      "pf = 0.9;         #Power Factor\n",
+      "Sm = 10360.;         #Load on Main Feeder in kVA\n",
+      "Vll = 12.47;         #Line to Line voltage in kV\n",
+      "Imain = Sm/(math.sqrt(3)*Vll);         #Current in Main Feeder\n",
+      "\n",
+      "#Note Suffix l means lateral and m means main\n",
+      "\n",
+      "Vph = 7.2;         #Phase Voltage in kV\n",
+      "Sl = 2*518.;         #Load on Lateral Feeder in kVA\n",
+      "Ilateral = Sl/Vph;         #Current in Laterals\n",
+      "\n",
+      "#Length of the Feeder\n",
+      "#Length Consmath.tant\n",
+      "Cm = 5280.;         #Main\n",
+      "Cl = 1000.;         #Lateral\n",
+      "Ll = 5760./Cl;         #Lateral Length\n",
+      "Lm = 3300./Cm;         #Main Length\n",
+      "\n",
+      "#Consmath.tants for the particular cables from the tables\n",
+      "rl = 0.331;\n",
+      "xLl = 0.0300;\n",
+      "rm = 0.342;\n",
+      "xam = 0.458;\n",
+      "xdm = 0.1802;\n",
+      "xLm = xam+xdm;\n",
+      "\n",
+      "# Calculations\n",
+      "#Voltage Drops for Normal Condition\n",
+      "VDmainn = (Imain/2)*((rm*pf)+(xLm*math.sin(math.radians(math.acos(pf)))))*Lm/2;\n",
+      "VDlateraln = (Ilateral/2)*((rl*pf)+(xLl*math.sin(math.radians(math.acos(pf)))))*Ll/2;\n",
+      "\n",
+      "perVDmainn = VDmainn*100/Vb;\n",
+      "perVDlateraln = VDlateraln*100/Vb;\n",
+      "\n",
+      "TVDn = perVDmainn+perVDlateraln;\n",
+      "\n",
+      "#Voltage Drops for Worst Conditions\n",
+      "VDmainw = (Imain)*((rm*pf)+(xLm*math.sin(math.radians(math.acos(pf)))))*Lm/2;\n",
+      "VDlateralw = (Ilateral)*((rl*pf)+(xLl*math.sin(math.radians(math.acos(pf)))))*Ll;\n",
+      "\n",
+      "perVDmainw = VDmainw*100/Vb;\n",
+      "perVDlateralw = VDlateralw*100/Vb;\n",
+      "\n",
+      "TVDw = perVDmainw+perVDlateralw;\n",
+      "\n",
+      "# Results\n",
+      "print 'a)From Table A5, 300-kcmilACSR conductors, with 500A Ampacity is used for mainand AWG         #2 XLPE Al\\\n",
+      " URD cable with 168A Ampacity'\n",
+      "print 'b) The Total Voltage Drop in Percent for Normal Operation is %g percent'%(TVDn)\n",
+      "print 'c) The Total Voltage Drop in Percent for Worst Condition is %g percent'%(TVDw)\n",
+      "print 'd The Voltage drop is met for Normal operation and NOT for emergency operation'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)From Table A5, 300-kcmilACSR conductors, with 500A Ampacity is used for mainand AWG         #2 XLPE Al URD cable with 168A Ampacity\n",
+        "b) The Total Voltage Drop in Percent for Normal Operation is 1.1836 percent\n",
+        "c) The Total Voltage Drop in Percent for Worst Condition is 4.08313 percent\n",
+        "d The Voltage drop is met for Normal operation and NOT for emergency operation\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch6_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch6_1.ipynb
new file mode 100755
index 00000000..811db39e
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch6_1.ipynb
@@ -0,0 +1,355 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:753590086967c8fe303e9fb46c959fc704437382516c05f043082b94c3ead7f2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Design Considerations of Secondary Systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1  Page No : 296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy import Symbol,solve\n",
+      "from scipy.misc import derivative\n",
+      "\n",
+      "# Variables\n",
+      "NC = 24.;         #Number Of Customers Per Block\n",
+      "\n",
+      "#We get the Total Annual Cost from the releveant equations as\n",
+      "# TAC  =  239.32 + (3.1805*ST) + (3492/ST) + (28170/ST**2) + (0.405*ASL) + (17018/ASL) + (1.134*ASD) + (8273/ASD)\n",
+      "\n",
+      "#We know split the above equation into 3 different parts according to factors ST,ASD,ASL\n",
+      "\n",
+      "#Variable Values of the Factors\n",
+      "ST = Symbol('ST');\n",
+      "ASD = Symbol('ASD');\n",
+      "ASL = Symbol('ASL');\n",
+      "\n",
+      "#Functions to Find the TAC corresponding to the Respective Factors\n",
+      "\n",
+      "def TransSize(y): \n",
+      "    return 239.52 + (3.1805*y) + (3492./y) + (28170./(y**2))\n",
+      "def SDwire(y): \n",
+      "    return (1.134*y)+(8273./y)\n",
+      "def SLwire(y): \n",
+      "    return (0.405*y)+(17018./y)\n",
+      "\n",
+      "#Total Annual Costs of the respective Factors\n",
+      "TACST  =  TransSize(ST);\n",
+      "TACASD  =  SDwire(ASD);\n",
+      "TACASL  =  SLwire(ASL);\n",
+      "\n",
+      "#Partially Differentiating wrt ASD we get\n",
+      "#Y1 = derivative(TACASD);\n",
+      "#X1 = roots(Y1[1]);\n",
+      "X1 = [-85.413198,85.413198]\n",
+      "ASD = X1[0];          #Calculated Value\n",
+      "ASDstd  =  105.500;\n",
+      "ASDstd1  =  133.1;\n",
+      "\n",
+      "#Partially Differentiating wrt ASL we get\n",
+      "#Y2 = derivative(TACASL);\n",
+      "#X2 = solve(Y2[1]);\n",
+      "X2 = [-204.9872,204.9872  ]\n",
+      " \n",
+      "ASL = X2[0];         #Calculated Value\n",
+      "ASLstd  =  211.600; \n",
+      "ASLstd1  =  250.;\n",
+      "\n",
+      "#Partially Differentiating wrt ST we get\n",
+      "#Y3 = derivative(TACST);\n",
+      "#X3 = solve(Y3[1]);\n",
+      "X3 = [39.346541,-19.673271 + 7.9480914j, -19.673271 - 7.9480914j,  0 ]\n",
+      "ST = round(X3[0]);          #Calculated Value\n",
+      "STstd  =  50.;\n",
+      "\n",
+      "#Total Annual Cost of the Calculated parameters\n",
+      "TAC = TransSize(ST)+SDwire(ASD)+SLwire(ASL);\n",
+      "#Calculation Mistake in The Text Book\n",
+      "\n",
+      "#Total Annual Cost of the First Higher standard Parameters\n",
+      "TACstd = TransSize(STstd)+SDwire(ASDstd)+SLwire(ASLstd);\n",
+      "#Total Annual Cost of the Second Higher standard Parameters\n",
+      "TACstd1 = TransSize(STstd)+SDwire(ASDstd1)+SLwire(ASLstd1);\n",
+      "\n",
+      "#Total Fixed Charges per Year\n",
+      "TACFC = ((75+(2.178*STstd))+(5.4+(0.405*ASLstd))+(15.12+(1.134*ASD))+(144));\n",
+      "#Total Operating Charges per Year\n",
+      "TACOC = ((0.0225*STstd)+(0.98*STstd)+(28170/(STstd**2))+(3492/STstd)+(17018/ASLstd)+(8273/ASDstd));\n",
+      "\n",
+      "#Values Might Vary from those in the text due to high precision\n",
+      "\n",
+      "#Fixed Charges Per Customer Per Month\n",
+      "FC = TACFC/(NC*12);\n",
+      "\n",
+      "#Variable Costs Per Customer per month\n",
+      "VOC = TACOC/(NC*12);\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Most Economical SD Size is %g kmil and the nearest larger standard AWG wire size is %g kmil'%(ASD,ASDstd)\n",
+      "print 'b) The Most Economical SL Size is %g kmil and the nearest larger standard AWG wire size is %g kmil'%(ASL,ASLstd)\n",
+      "print 'c) The Most Economical Distribution Transformer Size is %g kmil and the nearest larger standard transformer\\\n",
+      " size is %g kVA'%(ST,STstd)\n",
+      "print 'd) The Total Annual Cost Per Block for the theoretically most economical sizes of equipment is %g dollars'%(TAC)\n",
+      "print 'e) The Total Annual Cost Per Block for the nearest larger standard comercial sizes of equipment is %g dollars'%(TACstd)\n",
+      "print 'f) The Total Annual Cost Per Block for the nearest larger transformer size and for\\\n",
+      " the second larger sizes of ASD and ASL is %g dollars'%(TACstd1)\n",
+      "print 'g) Fixed Charges per Customer per Month is %g dollars'%(FC)\n",
+      "print 'h) The Variable Operating Costs Per Customer Per Month is %g dollars'%(VOC)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Most Economical SD Size is -85.4132 kmil and the nearest larger standard AWG wire size is 105.5 kmil\n",
+        "b) The Most Economical SL Size is -204.987 kmil and the nearest larger standard AWG wire size is 211.6 kmil\n",
+        "c) The Most Economical Distribution Transformer Size is 39 kmil and the nearest larger standard transformer size is 50 kVA\n",
+        "d) The Total Annual Cost Per Block for the theoretically most economical sizes of equipment is 111.862 dollars\n",
+        "e) The Total Annual Cost Per Block for the nearest larger standard comercial sizes of equipment is 843.83 dollars\n",
+        "f) The Total Annual Cost Per Block for the nearest larger transformer size and for the second larger sizes of ASD and ASL is 862.067 dollars\n",
+        "g) Fixed Charges per Customer per Month is 1.17104 dollars\n",
+        "h) The Variable Operating Costs Per Customer Per Month is 1.00721 dollars\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2  Page No : 304"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import matrix,multiply\n",
+      "#To determine the co-effcient matrix for a unbalanced load\n",
+      "#Page 304\n",
+      "\n",
+      "# Variables\n",
+      "Dab = 12.;\n",
+      "Dan = 12.;\n",
+      "Dbn = 24.;\n",
+      "Daa = 12*0.01577;\n",
+      "Dbb = Daa;\n",
+      "Dnn = Daa;\n",
+      "\n",
+      "# Calculations\n",
+      "def Coeff(y,z):\n",
+      "    return (2*(10**-7))*math.log(y/z)         #function to find the elements of the co-efficient matrix\n",
+      "\n",
+      "#Part A of the question cannot be found using Scilab, Hence from the equation obtained in part A we can numerically compute the Co- Efficient Matrix\n",
+      "\n",
+      "CM = matrix([[Coeff(Dan,Daa),Coeff(Dan,Dab)],[Coeff(Dbn,Dab),Coeff(Dbn,Dbb)],[Coeff(Dnn,Dan),Coeff(Dnn,Dbn)]]);\n",
+      "\n",
+      "# Results\n",
+      "print ' Part A cannot be resulted by this code%( hence from the equations obtained in Part A Co-Efficient Matrix is Obtained as'\n",
+      "print (multiply(CM,(10**7)))\n",
+      "print ' * 10**-7 Wb*T/m'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part A cannot be resulted by this code%( hence from the equations obtained in Part A Co-Efficient Matrix is Obtained as\n",
+        "[[ 8.29929176  0.        ]\n",
+        " [ 1.38629436  9.68558612]\n",
+        " [-8.29929176 -9.68558612]]\n",
+        " * 10**-7 Wb*T/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4  Page No : 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import matrix\n",
+      "from numpy.linalg import inv\n",
+      "\n",
+      "# Variables\n",
+      "#Primary Voltage\n",
+      "V1 = 7272*(1j*math.pi*0/180);\n",
+      "\n",
+      "#Secondary Voltages\n",
+      "Ea = 120*(1j*math.pi*0/180);\n",
+      "Eb = 120*(1j*math.pi*0/180);\n",
+      "\n",
+      "#Impedances\n",
+      "Za = 0.8+(1j*0.6);\n",
+      "Zb = 0.8+(1j*0.6);\n",
+      "\n",
+      "n = 60;         #Turns Ratio\n",
+      "\n",
+      "# Calculations\n",
+      "def angle(y): \n",
+      "    return math.degrees(math.atan(y.imag/y.real))\n",
+      "\n",
+      "#Substituting the values we get the following equations\n",
+      "#121.2  =  Ia*(0.8857 + j0.6846) + Ib*(0.03279 + j0.03899)\n",
+      "#121.2  =  Ia*(-0.03279 - j0.03899) + Ib*(-0.88574 + j0.50267)\n",
+      "\n",
+      "#For Convenience We segregate them as\n",
+      "Z1 = (0.8857+(1j*0.6846));\n",
+      "Z2 = (0.03279+(1j*.03899))\n",
+      "Z3 = (-0.03279-(1j*.03899))\n",
+      "Z4 = (-0.88574+(1j*.50267))\n",
+      "\n",
+      "Z = matrix([[Z1,Z2],[Z3,Z4]]);         #Impedance matrix\n",
+      "V = matrix([[121.2],[121.2]]);         #Voltage Matrix\n",
+      "I = inv(Z)*V;         #Current Matrix\n",
+      "\n",
+      "#Secondary Currents\n",
+      "Ia = I[0];\n",
+      "Ib = I[1];\n",
+      "\n",
+      "In = -Ia-Ib;         #Secondary neutral Currents\n",
+      "\n",
+      "#Secondary Voltages\n",
+      "Va = Za*Ia;\n",
+      "Vb = -1*Zb*Ib;\n",
+      "\n",
+      "#Secondary Voltage Resulmath.tant\n",
+      "Vab = Va+Vb;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Secondary Currents are:'\n",
+      "print 'Ia  =  %g/_%g A'%(abs(Ia),angle(Ia))\n",
+      "print 'Ib  =  %g/_%g A'%(abs(Ib),180+angle(Ib))\n",
+      "print 'b) The Secondary Neutral Current  =  %g/_%g A'%(abs(In),angle(In))\n",
+      "print 'c The Secondary Voltages are:'\n",
+      "print 'Va  =  %g/_%g V'%(abs(Va),angle(Va))\n",
+      "print 'Vb  =  %g/_%g V'%(abs(Vb),angle(Vb))\n",
+      "print 'd) The Resulmath.tant Secondary Voltage Vab is %g/_%g V'%(abs(Vab),angle(Vab))\n",
+      "\n",
+      "#In the TextBook Note That Zb has been taken wrong in the calculattion of Vb\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Secondary Currents are:\n",
+        "Ia  =  109.376/_-34.7812 A\n",
+        "Ib  =  124.295/_210.235 A\n",
+        "b) The Secondary Neutral Current  =  126.208/_82.0057 A\n",
+        "c The Secondary Voltages are:\n",
+        "Va  =  109.376/_2.08869 V\n",
+        "Vb  =  124.295/_67.1048 V\n",
+        "d) The Resulmath.tant Secondary Voltage Vab is 197.222/_36.9266 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5  Page No : 310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "N = 19.;         #Number Transformers\n",
+      "St = 500.;         #Load on each transformer in kVA\n",
+      "L = 5096+(1j*3158);         #Load\n",
+      "Vlf = 114.;         #Favourable Voltage\n",
+      "Vlt = 111.;         #Tolerable Volatage\n",
+      "Vb = 125.;         #Base Voltage\n",
+      "\n",
+      "# Calculations\n",
+      "#Per Unit Tolerable and favourable voltages\n",
+      "puVlf = Vlf/Vb;\n",
+      "puVlt = Vlt/Vb;\n",
+      "\n",
+      "ZM = 0.181+(1j*0.115);         #The Positive Sequence Impedance\n",
+      "ZTi = 0.0086+(1j*0.0492);         #Transformer Impedance for 500kVA\n",
+      "ZT = 2*ZTi;         #Transformer Impedance for 1000kVA\n",
+      "\n",
+      "AAF = N*St/abs(L);         #Actual Application Factor\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Lowest favourable Voltage is %g pu and The Lowest tolerable voltage is %g pu'%(puVlf,puVlt)\n",
+      "print 'b) There Are No buses in Table 6-5%( for the first contingency outage which satisfy the necessary condition'\n",
+      "print 'c For Second Contingency Outage'\n",
+      "print '1) Less than Favourable Voltage are B,C,J,K,R and S'\n",
+      "print '2) Less than Tolerable Voltage are B,C,J,K.'\n",
+      "print 'd) ZM/ZT  =  %g and 1/2)*ZM/ZT  =  %g respectively.'%(abs(ZM)/abs(ZT),((1./2)*abs(ZM)/abs(ZT)))\n",
+      "print 'The Actual Application Factor is %g'%(AAF)\n",
+      "print 'Therefore the Design of this network is sufficient'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Lowest favourable Voltage is 0.912 pu and The Lowest tolerable voltage is 0.888 pu\n",
+        "b) There Are No buses in Table 6-5%( for the first contingency outage which satisfy the necessary condition\n",
+        "c For Second Contingency Outage\n",
+        "1) Less than Favourable Voltage are B,C,J,K,R and S\n",
+        "2) Less than Tolerable Voltage are B,C,J,K.\n",
+        "d) ZM/ZT  =  2.14675 and 1/2)*ZM/ZT  =  1.07338 respectively.\n",
+        "The Actual Application Factor is 1.58461\n",
+        "Therefore the Design of this network is sufficient\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch7_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch7_1.ipynb
new file mode 100755
index 00000000..29035765
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch7_1.ipynb
@@ -0,0 +1,751 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:01457b400cb26c0846fbd5ddd5063f407fdfab264a58fb8bca41c527415d193b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Voltage Drop and Power Loss Calculations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2  Page No : 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Vll = 416.;         #Voltage Line to Line\n",
+      "Vph = Vll/(math.sqrt(3));         #Phase Voltage and Base Voltage\n",
+      "#Load Currents\n",
+      "Ia = 30.;\n",
+      "Ib = 20.;\n",
+      "Ic = 50.;\n",
+      "\n",
+      "#Power Factors of the load\n",
+      "pfa = 1.;\n",
+      "pfb = 0.5;\n",
+      "pfc = 0.9;\n",
+      "\n",
+      "#Impedances of the Sections\n",
+      "ZA = 0.05+(1j*0.01);\n",
+      "ZAB = 0.1+(1j*0.02);\n",
+      "ZBC = 0.05+(1j*0.05);\n",
+      "#Impedance upto the point of load\n",
+      "ZB = ZA+ZAB;\n",
+      "ZC = ZB+ZBC;\n",
+      "\n",
+      "# Calculations\n",
+      "#Function to Calculate Voltage Drop\n",
+      "def VD(a,b,c):\n",
+      "    return a*((b.real*c)+(b.imag*math.sin(math.acos(c))))\n",
+      "\n",
+      "#Voltage Drops at A,B and C\n",
+      "VDA = VD(Ia,ZA,pfa);\n",
+      "VDB = VD(Ib,ZB,pfb);\n",
+      "VDC = VD(Ic,ZC,pfc);\n",
+      "\n",
+      "TVD = VDA+VDB+VDC;         #Total Voltage Drop\n",
+      "\n",
+      "TVDpu = TVD/Vph;         # In Per Unit\n",
+      "\n",
+      "def Real(y,z):\n",
+      "    return  Vph*y*z         #Function to Calculate Real Power\n",
+      "def Reactive(y,z):\n",
+      "    return  Vph*y*math.sin(math.acos(z))         #Funtion to Calculate the Reactive power\n",
+      "\n",
+      "#Real Powers\n",
+      "Pa = Real(Ia,pfa);\n",
+      "Pb = Real(Ib,pfb);\n",
+      "Pc = Real(Ic,pfc);\n",
+      "P = Pa+Pb+Pc;         #Total Real Power\n",
+      "\n",
+      "\n",
+      "#Reactive Powers\n",
+      "Qa = Reactive(Ia,pfa);\n",
+      "Qb = Reactive(Ib,pfb);\n",
+      "Qc = Reactive(Ic,pfc);\n",
+      "Q = Qa+Qb+Qc;         #Total Reactive Power\n",
+      "\n",
+      "S = math.sqrt((P**2)+(Q**2));         #Total output from the Transformer\n",
+      "PF = P/S;         #Load Power Factor\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Total Voltage drop is %g pu'%(TVDpu)\n",
+      "print 'b) The Real Power per Phase is %g kW'%(P/1000)\n",
+      "print 'c) The Reactive Power per Phase is %g kVAr'%(Q/1000)\n",
+      "print 'd) The Kilovoltampere output and load factor is %g kVA and %g lagging'%(S/1000,PF)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Total Voltage drop is 0.0593859 pu\n",
+        "b) The Real Power per Phase is 20.4151 kW\n",
+        "c) The Reactive Power per Phase is 9.39455 kVAr\n",
+        "d) The Kilovoltampere output and load factor is 22.473 kVA and 0.90843 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3  Page No : 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "pf = 0.9;         #Power Factor\n",
+      "Vb = 120.;         #Base Voltage\n",
+      "#From The Tables\n",
+      "r = 0.334;         #Resismath.tance per thousand feet\n",
+      "x = 0.0299;         #Reacmath.tance per thousand feet\n",
+      "K1 = 0.02613;         #Voltage Drop\n",
+      "\n",
+      "#Assumed Cable\n",
+      "I = 100.;         #Secodary line Current\n",
+      "Ls = 100.;         #Length of Secondary line in feet\n",
+      "\n",
+      "# Calculations\n",
+      "R = r*Ls/1000;         # Resismath.tance Value for a 100 feet Line\n",
+      "X = x*Ls/1000;         # Reacmath.tance Value for a 100 feet Line\n",
+      "\n",
+      "VD = I*((R*pf)+(X*math.sin(math.radians(math.acos(pf)))));         #Voltage Drop\n",
+      "VDpu = VD/Vb;         #Per unit value\n",
+      "\n",
+      "# Results\n",
+      "print 'The Cable Selected is of 100 feet, carrying 100A and cable size         #2 AWG'\n",
+      "print 'The Voltage drop for the above cable is %g pu V'%(VDpu)\n",
+      "print 'The Above Value is Close to the Value%g pu V) in the table given.'%(K1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Cable Selected is of 100 feet, carrying 100A and cable size         #2 AWG\n",
+        "The Voltage drop for the above cable is 0.0250696 pu V\n",
+        "The Above Value is Close to the Value0.02613 pu V) in the table given.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4  Page No : 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Sts = (10+(11*4.4));         #Load Selected on the transformer\n",
+      "V = 240.;         #Voltage\n",
+      "Sta = 50.;         #Available Unit\n",
+      "pf = 0.9;         #Load Power Factor\n",
+      "I = (Sts/V)/(Sta/V);\n",
+      "\n",
+      "VDT = I*((0.0107*pf)+(0.0139*math.sin(math.radians(math.acos(pf)))));\n",
+      "\n",
+      "SLload = 10.+(3*6);\n",
+      "\n",
+      "# Calculations\n",
+      "def VD(a,b,c):\n",
+      "    return a*b*c/(10**4)         #Function to find Voltage Drop Per unit\n",
+      "\n",
+      "VDSL = VD(0.0088,116.7,150);\n",
+      "VDSD = VD(0.01683,41.76,70);\n",
+      "\n",
+      "TVD = VDT+VDSL+VDSD;\n",
+      "\n",
+      "Is = 80.;\n",
+      "Smotor = Is*V/1000;\n",
+      "pf1 = 0.5;\n",
+      "VDIPT = ((0.0107*pf1)+(0.0139*math.sin(math.radians(math.acos(pf1)))))*(Smotor/Sta);\n",
+      "\n",
+      "VDIPSL = VD(0.00636,80,150);\n",
+      "VDIPSD = VD(0.01089,80,70);\n",
+      "TVDIP = VDIPT+VDIPSL+VDIPSD;\n",
+      "\n",
+      "VDSL1 = VD(0.00769,116.7,150);\n",
+      "VDSD1 = VD(0.0136,41.6,70);\n",
+      "TVD1 = VDT+VDSL1+VDSD1;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Voltage drops are:'\n",
+      "print 'Transformer is %g pu V'%(VDT)\n",
+      "print 'Secondary Lines is %g pu V'%(VDSL)\n",
+      "print 'Service Drops is %g pu V'%(VDSD)\n",
+      "print 'Total is %g pu V'%(TVD)\n",
+      "print 'The Above Value doesn''t meet the required criterion'\n",
+      "print 'b The Voltage dip for motor starting are:'\n",
+      "print 'Transformer is %g pu V'%(VDIPT)\n",
+      "print 'Secondary Lines is %g pu V'%(VDIPSL)\n",
+      "print 'Service Drops is %g pu V'%(VDIPSD)\n",
+      "print 'Total is %g pu V'%(TVDIP)\n",
+      "print 'The Above Value does meet the required criterion'\n",
+      "print 'C The Voltage drops after changing the conductors are:'\n",
+      "print 'Transformer is %g pu V'%(VDT)\n",
+      "print 'Secondary Lines is %g pu V'%(VDSL1)\n",
+      "print 'Service Drops is %g pu V'%(VDSD1)\n",
+      "print 'Total is %g pu V'%(TVD1)\n",
+      "print 'The Above Value doesn''t meet the required criterion'\n",
+      "print 'Thus 350 kcmilcable size for the secondary lines and        \\\n",
+      " #2/0 AWG cable size for service drops to meet the criteria'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Voltage drops are:\n",
+        "Transformer is 0.0113756 pu V\n",
+        "Secondary Lines is 0.0154044 pu V\n",
+        "Service Drops is 0.00491975 pu V\n",
+        "Total is 0.0316998 pu V\n",
+        "The Above Value doesnt meet the required criterion\n",
+        "b The Voltage dip for motor starting are:\n",
+        "Transformer is 0.00215195 pu V\n",
+        "Secondary Lines is 0.007632 pu V\n",
+        "Service Drops is 0.0060984 pu V\n",
+        "Total is 0.0158824 pu V\n",
+        "The Above Value does meet the required criterion\n",
+        "C The Voltage drops after changing the conductors are:\n",
+        "Transformer is 0.0113756 pu V\n",
+        "Secondary Lines is 0.0134613 pu V\n",
+        "Service Drops is 0.00396032 pu V\n",
+        "Total is 0.0287973 pu V\n",
+        "The Above Value doesnt meet the required criterion\n",
+        "Thus 350 kcmilcable size for the secondary lines and         #2/0 AWG cable size for service drops to meet the criteria\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5  Page No : 336"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Length in feet\n",
+      "Lsd = 100.;         #Service Drop Line\n",
+      "Lsl = 200.;         #Secondary Line\n",
+      "\n",
+      "SB = 75.;         #Transformer Capacity in kVA\n",
+      "pf = 0.9;         #Load Power Factor\n",
+      "\n",
+      "#From the Tables\n",
+      "ISL = 129.17;         #Secondary Line Current\n",
+      "ISD = 41.67;         #Service Drop Current\n",
+      "KSD = 0.01683;         #Service Drop Cable Consmath.tant\n",
+      "KSL = 0.0136;         #Secondary Cable Consmath.tant\n",
+      "\n",
+      "#for Transformer\n",
+      "R = 0.0101;         #Resismath.tance in per unit\n",
+      "X = 0.0143;         #Reamath.tance in per unit\n",
+      "\n",
+      "# Calculations\n",
+      "#From the Diagram\n",
+      "ST = (3.+2+8+6)+(5+6+7+4)+(6+7+8+10);         #Total Load on transformer\n",
+      "\n",
+      "STpu = ST/SB;         #In Per unit;\n",
+      "\n",
+      "#The Above value also corresponds to the Current as Well\n",
+      "\n",
+      "I = STpu;         #Current in Per Unit\n",
+      "\n",
+      "VDT = I*((R*pf)+(X*math.sin(math.radians(math.acos(pf)))));         #Voltage Drop in the Transformer\n",
+      "VDSL = KSL*(Lsl*ISL/(10**4));         #Secondary Line\n",
+      "VDSD = KSD*(Lsd*ISD/(10**4));         #Service Drop Line\n",
+      "\n",
+      "VD = VDT+VDSD+VDSL;         #Total Voltage Drop\n",
+      "\n",
+      "# Results\n",
+      "print 'a)The Load in transformer is %g kVA or %g pu'%(ST,STpu)\n",
+      "print 'b) The Total Voltage Drop is %g pu V'%(VD)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)The Load in transformer is 72 kVA or 0.96 pu\n",
+        "b) The Total Voltage Drop is 0.0509818 pu V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7  Page No : 340"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "An = 4.;         #Service Area\n",
+      "l = 1.;         #Length of 0a\n",
+      "#Voltages in kV \n",
+      "Vll = 13.2;         #Line to line\n",
+      "Vln = 7.62;         #Line to neutral\n",
+      "\n",
+      "#Peak Loading\n",
+      "Dp = 1000.;         #Peak Loading Intensity per sq.miles\n",
+      "Sl = 2000.;         #Lumped Load in kVA\n",
+      "\n",
+      "#Off Peak Loading\n",
+      "Dop = 333.;         #Loading intensity\n",
+      "\n",
+      "VB = 7620.;         #Base Voltage\n",
+      "\n",
+      "Vs = 1.025;         #Substation Voltages\n",
+      "\n",
+      "Sn = Dp*An;         #Load Connected to the square shaped service area\n",
+      "Sm = Sn+Sl;         #Total Load\n",
+      "\n",
+      "K = 0.0003;         #Cable Consmath.tant\n",
+      "\n",
+      "# Calculations\n",
+      "VD0a = K*Sm*l;         #Voltage Drop between substation and a\n",
+      "lab = 2;          #Dismath.tance from a to b\n",
+      "VDab = (K*Sn*lab/2)+(K*Sl*lab);        #Voltage drop from b to c\n",
+      "lbc = 2;         #Dismath.tance from b to c\n",
+      "VDbc = 3*K*Sl*lbc;         #Voltage drop from b to c         #Change in Consmath.tant\n",
+      "\n",
+      "I = Sl/(math.sqrt(3)*(0.947*Vll));\n",
+      "Ib = Sl/(math.sqrt(3)*(Vll));        #BAse Current\n",
+      "\n",
+      "MIpu = I/Ib;         #Per Unit Current\n",
+      "\n",
+      "Ztpu = complex(0,0.05);\n",
+      "pf = 0.9;         #Load Power Factor\n",
+      "\n",
+      "Ipu = MIpu*exp(1j*math.pi*math.acos(pf)/180);\n",
+      "\n",
+      "#The Voltage has been tapped up 5 percent\n",
+      "\n",
+      "puVDcd = (abs(Ipu)*((Ztpu.real*pf)+(Ztpu.imag*math.sin(math.radians(math.acos(pf))))))-0.05;\n",
+      "VDcd = puVDcd*100;\n",
+      "def volt(a,b):\n",
+      "    return  (a-(b/100.))         #funtion to find out voltages\n",
+      "\n",
+      "#per unit Voltages\n",
+      "puVa = volt(Vs,VD0a);\n",
+      "puVb = volt(puVa,VDab);\n",
+      "puVc = volt(puVb,VDbc);\n",
+      "puVd = volt(puVc,VDcd);\n",
+      "\n",
+      "#Voltages in V\n",
+      "Va = puVa*VB;\n",
+      "Vb = puVb*VB;\n",
+      "Vc = puVc*VB;\n",
+      "Vd = puVd*VB;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Percentage drops are'\n",
+      "print ' Substation to a is %g percent'%(VD0a);\n",
+      "print ' a to b is %g percent'%(VDab);\n",
+      "print ' b to c is %g percent'%(VDbc);\n",
+      "print ' c to d is %g percent'%(VDcd);\n",
+      "print 'b The Per unit voltages are:'\n",
+      "print ' Point a is %g pu V'%(puVa)\n",
+      "print ' Point b is %g pu V'%(puVb)\n",
+      "print ' Point c is %g pu V'%(puVc)\n",
+      "print ' Point d is %g pu V'%(puVd)\n",
+      "print 'c The Line to Neutral voltages are:'\n",
+      "print ' Point a is %g  V'%(Va)\n",
+      "print ' Point b is %g  V'%(Vb)\n",
+      "print ' Point c is %g  V'%(Vc)\n",
+      "print ' Point d is %g  V'%(Vd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Percentage drops are\n",
+        " Substation to a is 1.8 percent\n",
+        " a to b is 2.4 percent\n",
+        " b to c is 3.6 percent\n",
+        " c to d is -4.95844 percent\n",
+        "b The Per unit voltages are:\n",
+        " Point a is 1.007 pu V\n",
+        " Point b is 0.983 pu V\n",
+        " Point c is 0.947 pu V\n",
+        " Point d is 0.996584 pu V\n",
+        "c The Line to Neutral voltages are:\n",
+        " Point a is 7673.34  V\n",
+        " Point b is 7490.46  V\n",
+        " Point c is 7216.14  V\n",
+        " Point d is 7593.97  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8  Page No : 340"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To determine the percent drop from the substation to various points\n",
+      "#Page 340\n",
+      "\n",
+      "# Variables\n",
+      "V0op = 1.;         #At off Peak\n",
+      "An = 4.;         #Service Area\n",
+      "l = 1.;         #Length of 0a\n",
+      "#Voltages in kV \n",
+      "Vll = 13.2;         #Line to line\n",
+      "Vln = 7.62;         #Line to neutral\n",
+      "\n",
+      "#Peak Loading\n",
+      "Dp = 1000.;         #Peak Loading Intensity per sq.miles\n",
+      "Sl = 2000.;         #Lumped Load in kVA\n",
+      "\n",
+      "#Off Peak Loading\n",
+      "Dop = 333.;         #Loading intensity\n",
+      "\n",
+      "VB = 7620.;         #Base Voltage\n",
+      "\n",
+      "Vs = 1.025;         #Substation Voltages\n",
+      "\n",
+      "Sn = Dop*An;         #Load Connected to the square shaped service area\n",
+      "Sm = Sn+Sl;         #Total Load\n",
+      "\n",
+      "K = 0.0003;         #Cable Consmath.tant\n",
+      "\n",
+      "# Calculations\n",
+      "VD0a = K*Sm*l;         #Voltage Drop between substation and a\n",
+      "lab = 2.;          #Dismath.tance from a to b\n",
+      "VDab = (K*Sn*lab/2)+(K*Sl*lab);        #Voltage drop from b to c\n",
+      "lbc = 2.;         #Dismath.tance from b to c\n",
+      "VDbc = 3*K*Sl*lbc;         #Voltage drop from b to c         #Change in Consmath.tant\n",
+      "\n",
+      "I = Sl/(math.sqrt(3)*(0.947*Vll));\n",
+      "Ib = Sl/(math.sqrt(3)*(Vll));        #BAse Current\n",
+      "\n",
+      "MIpu = I/Ib;         #Per Unit Current\n",
+      "\n",
+      "Ztpu = complex(0,0.05);\n",
+      "pf = 0.9;         #Load Power Factor\n",
+      "\n",
+      "Ipu = MIpu*exp(1j*math.pi*math.acos(pf)/180);\n",
+      "\n",
+      "#The Voltage has been tapped up 5 percent\n",
+      "\n",
+      "puVDcd = (abs(Ipu)*((Ztpu.real*pf)+(Ztpu.imag*math.sin(math.radians(math.acos(pf))))))-0.05;\n",
+      "VDcd = puVDcd*100;\n",
+      "def volt(a,b):\n",
+      "    return (a-(b/100))        #funtion to find out voltages\n",
+      "\n",
+      "#per unit Voltages\n",
+      "puVa = volt(V0op,VD0a);\n",
+      "puVb = volt(puVa,VDab);\n",
+      "puVc = volt(puVb,VDbc);\n",
+      "puVd = volt(puVc,VDcd);\n",
+      "\n",
+      "#Voltages in V\n",
+      "Va = puVa*VB;\n",
+      "Vb = puVb*VB;\n",
+      "Vc = puVc*VB;\n",
+      "Vd = puVd*VB;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Percentage drops are'\n",
+      "print ' Substation to a is %g percent'%(VD0a);\n",
+      "print ' a to b is %g percent'%(VDab);\n",
+      "print ' b to c is %g percent'%(VDbc);\n",
+      "print ' c to d is %g percent'%(VDcd);\n",
+      "print 'b The Per unit voltages are:'\n",
+      "print ' Point a is %g pu V'%(puVa)\n",
+      "print ' Point b is %g pu V'%(puVb)\n",
+      "print ' Point c is %g pu V'%(puVc)\n",
+      "print ' Point d is %g pu V'%(puVd)\n",
+      "print 'c The Line to Neutral voltages are:'\n",
+      "print ' Point a is %g  V'%(Va)\n",
+      "print ' Point b is %g  V'%(Vb)\n",
+      "print ' Point c is %g  V'%(Vc)\n",
+      "print ' Point d is %g  V'%(Vd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Percentage drops are\n",
+        " Substation to a is 0.9996 percent\n",
+        " a to b is 1.5996 percent\n",
+        " b to c is 3.6 percent\n",
+        " c to d is -4.95844 percent\n",
+        "b The Per unit voltages are:\n",
+        " Point a is 0.990004 pu V\n",
+        " Point b is 0.974008 pu V\n",
+        " Point c is 0.938008 pu V\n",
+        " Point d is 0.987592 pu V\n",
+        "c The Line to Neutral voltages are:\n",
+        " Point a is 7543.83  V\n",
+        " Point b is 7421.94  V\n",
+        " Point c is 7147.62  V\n",
+        " Point d is 7525.45  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9  Page No : 344"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Vll = 13.2;         #Voltage in kV (Line voltage)\n",
+      "TCDi = 0.45;         #Load Density in kVA per feet\n",
+      "FD = 1.08;         #Diversity Factor for all loads\n",
+      "FLS = 0.2;         #Annual Loss Factor\n",
+      "DFi = 0.6;         #30 min Annual Demand Factor\n",
+      "\n",
+      "Dg = TCDi*DFi/FD;         #Divesified Maximum Demand of the Group\n",
+      "\n",
+      "L = 30000;         #Length of the Whole Feeder in Feet\n",
+      "\n",
+      "#To Achieve Minimum Voltage Drop, The Substation must be located at the centre of the line \n",
+      "Ln = 15000.;         #NEW Length of the Feeder\n",
+      "\n",
+      "# Calculations\n",
+      "SPK = Dg*Ln;         #Peak Load on each main of the substation trnasformer\n",
+      "I = (SPK/(math.sqrt(3)*Vll));         #Current in the Line\n",
+      "\n",
+      "K = 0.0009;         #For the Assumed Conductor\n",
+      "VD = K*SPK*Ln/(2*5280);         #Voltage Drop, Is divided by 5280, to convert the length in miles\n",
+      "\n",
+      "# Results\n",
+      "print 'a) To Achieve Minimum Voltage Drop, The Substation is Placed at the Centre of the Line,\\\n",
+      " and For a Current of %g A following in it,         #4 AWG or         #2 AWG ACSR conductors are used'%(I)\n",
+      "print 'b) For a         #4 AWG Copper Conductor, The Percentage Voltage drop at annual peak load is %g pecent'%(VD)\n",
+      "\n",
+      "#Calculation Mistake in the TextBook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) To Achieve Minimum Voltage Drop, The Substation is Placed at the Centre of the Line, and For a Current of 164.02 A following in it,         #4 AWG or         #2 AWG ACSR conductors are used\n",
+        "b) For a         #4 AWG Copper Conductor, The Percentage Voltage drop at annual peak load is 4.79403 pecent\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10  Page No : 346"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "Vll = 13.2;         #Voltage in kV (Line voltage)\n",
+      "TCDi = 0.45;         #Load Density in kVA per feet\n",
+      "FD = 1.08;         #Diversity Factor for all loads\n",
+      "FLS = 0.2;         #Annual Loss Factor\n",
+      "DFi = 0.6;         #30 min Annual Demand Factor\n",
+      "\n",
+      "# Calculations\n",
+      "Dg = TCDi*DFi/FD;         #Divesified Maximum Demand of the Group\n",
+      "L = 30000;         #Length of the Whole Feeder in Feet\n",
+      "I = 164.2;         #Current\n",
+      "\n",
+      "r = 0.592;         #Resismath.tance Per Mile\n",
+      "R = r*L/(3.*5280);         #Total Resismath.tance\n",
+      "\n",
+      "CL = 3*(I**2)*R;         #Total Power Loss in entire length\n",
+      "\n",
+      "TAEL = CL*FLS*8760/(10**3);         #Total Annual Energy Loss\n",
+      "\n",
+      "# Results\n",
+      "print 'The Total Annual Eddy Current Loss is %g kWhr'%(TAEL)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Total Annual Eddy Current Loss is 158887 kWhr\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11  Page No : 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Loads in kVA\n",
+      "Sbc = 3000.;         #Load Along bc\n",
+      "Sd = 2000.;         #Load At Point d\n",
+      "S0a = Sbc+Sd;         #Total Load\n",
+      "Sab = Sbc/2;         #Load along ab\n",
+      "\n",
+      "#Cable Consmath.tants\n",
+      "K0a = 0.0005;         #For 0 to a\n",
+      "Kab = 0.0010;         #For a to b\n",
+      "Kac = 0.0010;         #For a to c\n",
+      "Kad = 0.0010;         #For a to d\n",
+      "\n",
+      "#Length\n",
+      "l0a = 1.0;         #From 0 to a\n",
+      "lab = 2.;        #From a to b\n",
+      "lad = 2.;         #From a to d\n",
+      "\n",
+      "# Calculations\n",
+      "#Voltage Drops At Specific Points\n",
+      "VDa = K0a*S0a*l0a;\n",
+      "VDb = (Kab*Sab*lab/2)+VDa;\n",
+      "VDc = VDb;\n",
+      "VDd = (Kad*Sd*lad)+VDa;\n",
+      "\n",
+      "#To determine the Voltages at Point a\n",
+      "Vll = 12650.;         #Line to Line Voltage\n",
+      "Vln = 7300.;         #Line to Neutral Voltage\n",
+      "\n",
+      "Valn = Vln-(VDa*Vln/100);\n",
+      "Vall = Vll-(VDa*Vll/100);\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Voltage Drops at:'\n",
+      "print 'Point a is %g percent'%(VDa)\n",
+      "print 'Point b is %g percent'%(VDb)\n",
+      "print 'Point c is %g percent'%(VDc)\n",
+      "print 'Point d is %g percent'%(VDd)\n",
+      "print 'b) The Voltages VaL-N is %g V and VaL-L is %g V'%(Valn,Vall)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Voltage Drops at:\n",
+        "Point a is 2.5 percent\n",
+        "Point b is 4 percent\n",
+        "Point c is 4 percent\n",
+        "Point d is 6.5 percent\n",
+        "b) The Voltages VaL-N is 7117.5 V and VaL-L is 12333.8 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch8_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch8_1.ipynb
new file mode 100755
index 00000000..af025a65
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch8_1.ipynb
@@ -0,0 +1,323 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:82754afbe0b57de92c1b3631997744ed83c0b5bcb3ad977be765225d89fd57a7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Application of Capacitors to Distribution Systems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1  Page No : 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "SL = 700.;         #Load in kVA\n",
+      "pf1 = 65./100;         #Power Factor\n",
+      "PL = SL*pf1;         #Real Power\n",
+      "#From the Table of Power Factor Correction\n",
+      "CR = 0.74;         #Co-relation factor\n",
+      "CS = PL*CR;         #Capacitor Size\n",
+      "\n",
+      "CSr = 360.;         #Next Higher standard Capacitor Size\n",
+      "\n",
+      "# Calculations\n",
+      "CRn = CSr/PL;         #New Co-Relation Factor\n",
+      "\n",
+      "#From the table by linear interpolation\n",
+      "pfr = 93.;         #In Percentage\n",
+      "pfn = pfr+(172./320);\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Correction Factor is %g'%(CR)\n",
+      "print 'b) The Capacitor Size Required is %g kVAr'%(CS)\n",
+      "print 'c) Resulting power factor if the next higher standard capacitor size is used is %g percent'%(pfn)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Correction Factor is 0.74\n",
+        "b) The Capacitor Size Required is 336.7 kVAr\n",
+        "c) Resulting power factor if the next higher standard capacitor size is used is 93.5375 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2  Page No : 393"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Vll = 4.16;         #Line to Line Voltage in kV\n",
+      "Pr = (500*0.7457);         #Rating of motor in kW\n",
+      "pf1 = 0.75;         #Initial Power Factor\n",
+      "pfn = 0.9;         #Improved Power Factor\n",
+      "eff = 0.88;         #Efficiency\n",
+      "P = Pr/eff;         #Input Power of Induction Motor\n",
+      "\n",
+      "# Calculations\n",
+      "Q1 = P*math.degrees(math.atan(math.acos(pf1)));         #Reactive Power\n",
+      "Q2 = P*math.degrees(math.atan(math.acos(pfn)));         #REactive power of motor after power factor improvement\n",
+      "f = 60.;         #Frequency of supply\n",
+      "w = 2.*math.pi*f;         #Angular Frequency\n",
+      "Qc = Q1-Q2;         #Reactive Power of Capacitor\n",
+      "Il = Qc/(math.sqrt(3)*Vll);\n",
+      "\n",
+      "#Capacitor Connectd in Delta\n",
+      "Ic1 = Il/(math.sqrt(3));\n",
+      "Xc1 = Vll*1000/Ic1;         #Reacmath.tance of each capacitor\n",
+      "C1 = (10**6)/(w*Xc1);         #Capacimath.tance in Micro Farad\n",
+      "\n",
+      "#Capacitor Connected in Wye\n",
+      "Ic2 = Il;\n",
+      "Xc2 = Vll*1000/(math.sqrt(3)*Ic2);         #Reacmath.tance of each capacitor\n",
+      "C2 = (10**6)/(w*Xc2);         #Capacimath.tance in Micro Farad\n",
+      "\n",
+      "# Results\n",
+      "print 'a) Rating of Capacitor Bank is %g kVAr'%(Qc)\n",
+      "print 'b) The Value of Capacimath.tance if there are connected in delta is %g micro F'%(C1)\n",
+      "print 'c) The Value of Capacimath.tance if there are connected in wye is %g micro F'%(C2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) Rating of Capacitor Bank is 4906.48 kVAr\n",
+        "b) The Value of Capacimath.tance if there are connected in delta is 250.687 micro F\n",
+        "c) The Value of Capacimath.tance if there are connected in wye is 752.06 micro F\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3  Page No : 396"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "V = 2.4;         #Voltage in kV\n",
+      "I = 200;         #Load Current\n",
+      "P = 360;         #Real Load in kW\n",
+      "S1 = V*I;         #Total Load in kVA\n",
+      "pf1 = P/S1;         #Power Factor\n",
+      "Q1 = S1*math.sin(math.radians(math.acos(pf1)));         #Reactive Load\n",
+      "\n",
+      "# Calculations\n",
+      "Qc = 300;         #Capacitor Size\n",
+      "\n",
+      "Q2 = Q1-Qc;         #The New Reactive Load\n",
+      "pf2 = P/math.sqrt((P**2)+((Q1-Qc)**2));         #Improved Power Factor\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Uncorrected power factor and reactive load is %g and %g kVAr'%(pf1,Q1)\n",
+      "print 'b) The New Corrected factor after the introduction of capacitor unit is %g'%(pf2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Uncorrected power factor and reactive load is 0.75 and 6.0546 kVAr\n",
+        "b) The New Corrected factor after the introduction of capacitor unit is 0.77459\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4  Page No : 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "S1 = 7800.;         #Peak Load in kVA\n",
+      "T = 3*2000.;         #Total Rating of the Transformer\n",
+      "pf1 = 0.89;         #Load Power Factor\n",
+      "TC = 120./100;         #Thermal Capability\n",
+      "Qc = 1000.;         #Size of capacitor\n",
+      "\n",
+      "# Calculations\n",
+      "P = S1*pf1;         #Real Load\n",
+      "Q1 = S1*math.sin(math.radians(math.acos(pf1)));         #Reactive Load\n",
+      "\n",
+      "Q2 = Q1-Qc;         #The New Reactive Load\n",
+      "pf2 = P/math.sqrt((P**2)+((Q1-Qc)**2));         #Improved Power Factor\n",
+      "\n",
+      "S2 = P/pf2;         #Corrected Apprarent power\n",
+      "\n",
+      "ST = T*TC;         #Transformer Capabilty\n",
+      "\n",
+      "pf3 = P/ST;         #New Corrected Power factor required\n",
+      "\n",
+      "Q2new = P*math.degrees(math.atan(math.acos(pf3)));         #Required Reactive Power\n",
+      "Qcadd = Q2-Q2new;         #Additional Rating of the Capacitor\n",
+      "\n",
+      "# Results\n",
+      "print 'a) Since the Apparent Power%g kVAr is more than Transformer Capability %g kVAr),\\\n",
+      " Hence Additional Capacitors are required'%(S2,ST)\n",
+      "print 'b) The Rating of the Addtional capacitor is %g kVAr'%(Qcadd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) Since the Apparent Power7004.76 kVAr is more than Transformer Capability 7200 kVAr), Hence Additional Capacitors are required\n",
+        "b) The Rating of the Addtional capacitor is -105274 kVAr\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5  Page No : 411"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array\n",
+      "\n",
+      "# Variables\n",
+      "# 1 is Total Loss Reduction due to Capacitors\n",
+      "# 2 is Additional Loss Reduction due to Capacitor\n",
+      "# 3 is Total Demand Reduction due to capacitor\n",
+      "# 4 is Total required capacitor additions\n",
+      "\n",
+      "C90 = array([495165,85771,22506007,9810141]);         #Characteristics at 90% Power Factor\n",
+      "C98 = array([491738,75343,21172616,4213297]);         #Characteristics at 98% Power Factor\n",
+      "\n",
+      "#Responsibility Factors\n",
+      "RF90 = 1;\n",
+      "RF98 = 0.9;\n",
+      "\n",
+      "SLF = 0.17;         #System Loss Factor\n",
+      "FCR = 0.2;         #Fixed Charge Rate\n",
+      "DC = 250;         #Demand Cost\n",
+      "ACC = 4.75;         #Average Capacitor Cost\n",
+      "EC = 0.045;         #Energy Cost\n",
+      "Cd = C90-C98;         #Difference in Characteristics\n",
+      "\n",
+      "# Calculations\n",
+      "TAS = Cd[0]+Cd[1];         #Total Additional Kilowatt Savings\n",
+      "\n",
+      "ASDR1 = Cd[0]*RF90*DC*FCR;\n",
+      "ASDR2 = Cd[1]*RF98*DC*FCR;\n",
+      "TASDR = ASDR1+ASDR2;         #Total Annual Savings due to demand\n",
+      "x = 27;         # Cost for Per kVA of the capacity\n",
+      "TASTC = Cd[2]*FCR*x;         #Annual Savings due to Transmission Capacity\n",
+      "TASEL = TAS*SLF*EC*8760;         #Savings due to energy loss reduction\n",
+      "TACAC = Cd[3]*FCR*ACC;         #Annual Cost of Additional Capacitors\n",
+      "Savings = TASEL+TASDR+TASTC;         #Total Savings\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Resulting additional savings in kilowatt losses due to power factor improvement at the\\\n",
+      " substation buses is %g kW'%(Cd[0])\n",
+      "print 'b) The Resulting assitional savings in kilowatt  losses due to the power factor improvement for feeders is %g kW'%(Cd[1])\n",
+      "print 'c) The Additional Kilowatt Savings is %g kW'%(TAS)\n",
+      "print 'd) The Additional savings in the system kilovoltampere capacity is %g kVA'%(Cd[2])\n",
+      "print 'e) The Additional Capacitors required are %g kVAr'%(Cd[3])\n",
+      "print 'f) The Annual Savings in demand reduction due to capacitors applied to distribution substation buses\\\n",
+      " is approximately is %g dollars/year'%(TASDR)\n",
+      "print 'g) The Annual Savings due to the additional released transmission capacity is %g dollars/year'%(TASTC)\n",
+      "print 'h) The Total Anuual Savings due to the energy loss reduction is %g dollars/year'%(TASEL)\n",
+      "print 'i) The Total Annual Cost of the additional capacitors is %g dollars/year'%(TACAC)\n",
+      "print 'j) The Total Annual Savings is %g dollars/year'%(Savings)\n",
+      "print 'k) No Since the total net annual savings is not zero'\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Resulting additional savings in kilowatt losses due to power factor improvement at the substation buses is 3427 kW\n",
+        "b) The Resulting assitional savings in kilowatt  losses due to the power factor improvement for feeders is 10428 kW\n",
+        "c) The Additional Kilowatt Savings is 13855 kW\n",
+        "d) The Additional savings in the system kilovoltampere capacity is 1.33339e+06 kVA\n",
+        "e) The Additional Capacitors required are 5.59684e+06 kVAr\n",
+        "f) The Annual Savings in demand reduction due to capacitors applied to distribution substation buses is approximately is 640610 dollars/year\n",
+        "g) The Annual Savings due to the additional released transmission capacity is 7.20031e+06 dollars/year\n",
+        "h) The Total Anuual Savings due to the energy loss reduction is 928479 dollars/year\n",
+        "i) The Total Annual Cost of the additional capacitors is 5.317e+06 dollars/year\n",
+        "j) The Total Annual Savings is 8.7694e+06 dollars/year\n",
+        "k) No Since the total net annual savings is not zero\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch9_1.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch9_1.ipynb
new file mode 100755
index 00000000..ecdc82d7
--- /dev/null
+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch9_1.ipynb
@@ -0,0 +1,715 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:34811960951969c03d923c2b2288130419dca2ae4cd032dc1740e25d2d5fd68b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Distribution System Voltage Regulation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1  Page No : 468"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "#Base Value\n",
+      "S3phib = 15;         #in MVA\n",
+      "Vllst = 69;         #in kV\n",
+      "Vllp = 13.2;         #in kV\n",
+      "Vrrb = 120.;\n",
+      "\n",
+      "Ztpu = 1j*0.08;         #Transformer impedance per unit length\n",
+      "VSTpuop = 1.05*exp(1j*0);         #Per Unit Maximum Subtransmission Voltage Off Peak\n",
+      "VSTpup = 1.00*exp(1j*0);         #Per Unit Maximum Subtransmission Voltage Peak\n",
+      "pftop = 0.95;         #Off Peak kilovoltageamperage power factor\n",
+      "Sop = 0.25;         #Off Peak kilovoltageamperage\n",
+      "pftp = 0.85;         #Off Peak kilovoltageamperage power factor\n",
+      "Sp = 1.0;         #Off Peak kilovoltageamperage\n",
+      "Regpu = 5./(8*100);         #Regulated percent volts for 32 steps\n",
+      "K = 3.88*(10**-6);         #Drop Consmath.tant\n",
+      "S = 4000.;         # Peak Load in kVA\n",
+      "l = 10.;         #Length of the feeder\n",
+      "#Case A\n",
+      "Rset = 0.;\n",
+      "Xset = 0.;\n",
+      "Vpmax = 1.0417;\n",
+      "BW = 0.0083;\n",
+      "VRRpu = (Vpmax-BW);         #Setting of VRR\n",
+      "VRR = (Vpmax-BW)*Vrrb;\n",
+      "\n",
+      "# Calculations\n",
+      "#Case B\n",
+      "IPpuop = (Sop/VSTpuop)*exp(1j*math.acos(pftop)*math.pi/180);         #No Load Primary Current at substation Off Peak\n",
+      "VPpuop  = VSTpuop-(IPpuop*Ztpu);         #Highest Allowable Primary Voltage Off Peak\n",
+      "IPpup = (Sp/VSTpup)*exp(-1*1j*math.acos(pftp)*math.pi/180);         #No Load Primary Current at substation Peak\n",
+      "VPpup  = VSTpup-(IPpup*Ztpu);         #Highest Allowable Primary Voltage  Peak\n",
+      "\n",
+      "Step1 = (abs(VPpuop)-VRRpu)/Regpu;         #Off Peak Number Steps\n",
+      "#To find the positive step value\n",
+      "Step2 = -1*(abs(VPpup)-VRRpu)/Regpu;         # Peak Number Steps\n",
+      "\n",
+      "\n",
+      "#Case C\n",
+      "#At Peak Load Primary Voltages\n",
+      "MaxVpp = 1.075;         #Max\n",
+      "MinVpp = 1.000;         #Min\n",
+      "\n",
+      "TVDpu = K*S*l/2;         #Total Voltage Drop\n",
+      "MinVPpu = VRRpu-TVDpu;\n",
+      "\n",
+      "#At Annual Peak Load Primary Voltages\n",
+      "APMaxVPpu = MaxVpp-BW;         #Max\n",
+      "APMinVPpu = MinVpp+BW;         #Min\n",
+      "\n",
+      "#At no load Load Primary Voltages\n",
+      "NLMaxVPpu = Vpmax-BW;         #Max\n",
+      "NLMinVPpu = APMinVPpu;         #Min\n",
+      "\n",
+      "# Results\n",
+      "print 'a)The Setting of the VRR for the highest allowable primary voltage is %g V'%(VRR)\n",
+      "print 'b The Maximum Number of Steps of buck and boost for:'\n",
+      "print 'Off Peak : %g steps'%(math.ceil(Step1))\n",
+      "print 'Peak : %g steps'%(math.ceil(Step2))\n",
+      "print 'c) At Annual Load%( Significant Values on Voltage Curve'\n",
+      "print 'The Total Voltage Drop is %g pu V'%(TVDpu)\n",
+      "print 'The Minimum Feeder Voltage at the end of the feeder is %g'%(MinVPpu)\n",
+      "print 'The Minimum and Maximum Primary Voltages at Peak Load is %g pu V and %g pu V'%(APMaxVPpu,APMinVPpu)\n",
+      "print 'The Minimum and Maximum Primary Voltages at Peak Load is %g pu V and %g pu V'%(NLMaxVPpu,NLMinVPpu)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)The Setting of the VRR for the highest allowable primary voltage is 124.008 V\n",
+        "b The Maximum Number of Steps of buck and boost for:\n",
+        "Off Peak : 3 steps\n",
+        "Peak : 5 steps\n",
+        "c) At Annual Load%( Significant Values on Voltage Curve\n",
+        "The Total Voltage Drop is 0.0776 pu V\n",
+        "The Minimum Feeder Voltage at the end of the feeder is 0.9558\n",
+        "The Minimum and Maximum Primary Voltages at Peak Load is 1.0667 pu V and 1.0083 pu V\n",
+        "The Minimum and Maximum Primary Voltages at Peak Load is 1.0334 pu V and 1.0083 pu V\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2  Page No : 472"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy import Symbol,solve\n",
+      "\n",
+      "# Variables\n",
+      "#Terms from previous example\n",
+      "TVDpu = 0.0776;         #Total Voltage Drop\n",
+      "VRRpu = 1.035;         #Setting Voltage of Regulator\n",
+      "l = 10.;        #Length of the Feeder\n",
+      "\n",
+      "#Primary voltages for various cases\n",
+      "VPpua = 1.01;\n",
+      "VPpub = 1.00;\n",
+      "\n",
+      "s1 = Symbol('s1');         #Variable Value of Regulator length\n",
+      "#Function to find the equation for the regulator dismath.tance\n",
+      "def dist(y): \n",
+      "    return (s1*(2-(s1/l))/l)-((VRRpu-y)/TVDpu)\n",
+      "\n",
+      "# Calculations\n",
+      "#Different Cases\n",
+      "Xa = dist(VPpua);\n",
+      "Xb = dist(VPpub);\n",
+      "\n",
+      "s1a = solve(Xa);\n",
+      "if((abs(l-s1a[0])+(l-s1a[0])) == 0):\n",
+      "    s1a = s1a[1];\n",
+      "else:\n",
+      "    s1a = s1a[0];\n",
+      "\n",
+      "s1b = solve(Xb);\n",
+      "if((abs(l-s1b[0])+(l-s1b[0])) == 0):\n",
+      "    s1b = s1b[1];\n",
+      "else:\n",
+      "    s1b = s1b[0];\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Regulator must be placed at %g miles from the start of the feeder'%(s1a)\n",
+      "print 'b) The Regulator must be placed at %g miles from the start of the feeder'%(s1b)\n",
+      "print 'c The Advantage of a over b is that it can compensate for future growth'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Regulator must be placed at 1.76693 miles from the start of the feeder\n",
+        "b) The Regulator must be placed at 2.59076 miles from the start of the feeder\n",
+        "c The Advantage of a over b is that it can compensate for future growth\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3  Page No : 473"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "l = 10.;         #Length of the feeder\n",
+      "S3phi = 4000.;         #Annual Peak Load in kVA\n",
+      "VPpu = 1.01;         #Primary Feeder Voltage\n",
+      "s1 = 1.75;         # Dismath.tance of the Regulator\n",
+      "Rmax = 10./100;         #Regulation Percent\n",
+      "\n",
+      "# Calculations\n",
+      "S = S3phi*(1-(s1/l));         #Uniformly Distributed three phase load\n",
+      "Sph = S/3;         #Single Phase Load\n",
+      "\n",
+      "Sreg = Rmax*Sph;         #Regulated Size\n",
+      "\n",
+      "# Results\n",
+      "print 'The Calculated Circuit Kilovoltampere Size is %g kVA, And The corresponding Minimum kilovoltampere size \\\n",
+      "of the regulator size can be found as 114.3 kVA'%(Sreg)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Calculated Circuit Kilovoltampere Size is 110 kVA, And The corresponding Minimum kilovoltampere size of the regulator size can be found as 114.3 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4  Page No : 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To specify the best settings for regulation\n",
+      "#Page 474\n",
+      "\n",
+      "# Variables\n",
+      "s1 = 1.75;         #As Found in Example 2\n",
+      "VRRpu = 1.035;         #As R and X are zero, the Settings turn out to produce this\n",
+      "\n",
+      "#Parameters of Distribution\n",
+      "K = 3.88*(10**-6);\n",
+      "S = 3300.;\n",
+      "l = 10.;         #length of the line\n",
+      "\n",
+      "# Calculations\n",
+      "VDpu = K*S*(l-s1)/2;         #Per unit voltage drop\n",
+      "\n",
+      "VP = VRRpu-VDpu;         #Primary Feeder Voltage\n",
+      "\n",
+      "#We Obtain VDs  =  K*(S3-((S3*s)/l))s+K(S*s/l)s/2;\n",
+      "#We take various values of s and carry out the computation and hence form the table 9-4 given given in the result file\n",
+      "\n",
+      "#We Obtain from the voltage drop value for any give point s between the substation and the regulator  station as\n",
+      "#VDs = I(r.math.cos(theta)+ del math.sin(theta))s*(1-(s/(2*l)))\n",
+      "\n",
+      "#We finally Get VDs  =  3.88 * (10**-6) * (3300-(3300s/8.25))s+3.88*(10**-6)*(3300s/8.25)*s/2\n",
+      "\n",
+      "# Results\n",
+      "#Again for different values of s we get the table 9-5\n",
+      "print 'a)The Best Settings for LDCs R and X, and for the VRR'\n",
+      "print 'The best settings for LDC of the regulator are when settings for both R and X are zero and VRRpu  =  %g pu V'%(VRRpu)\n",
+      "print 'b)The Voltage Drop occuring in the feeder portion between the regulating point and the end of the feeder is %g pu V'%(VDpu)\n",
+      "print 'The Result Files give the Profiles and relevant information about the solution'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)The Best Settings for LDCs R and X, and for the VRR\n",
+        "The best settings for LDC of the regulator are when settings for both R and X are zero and VRRpu  =  1.035 pu V\n",
+        "b)The Voltage Drop occuring in the feeder portion between the regulating point and the end of the feeder is 0.0528165 pu V\n",
+        "The Result Files give the Profiles and relevant information about the solution\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5  Page No : 478"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To determine the setting of the regulator so that the voltage criteria is met\n",
+      "#Page 478\n",
+      "\n",
+      "# Variables\n",
+      "l = 10.;         #Length of the feeder\n",
+      "s1 = 1.75;\n",
+      "ra = 0.386;\n",
+      "xa = 0.4809;\n",
+      "xd = 0.1802;\n",
+      "xL = xa+xd;\n",
+      "Vb = 120;\n",
+      "pf = 0.85;         #Power Factor\n",
+      "S = 1100.;         #Load in kVA\n",
+      "Vln = 7.62;         #line to neutral voltage in kV\n",
+      "Reff = ra*(l-s1)/2;\n",
+      "Xeff = xL*(l-s1)/2;\n",
+      "\n",
+      "#Primary Ratings\n",
+      "CTp = 150;         #Current Tranformer\n",
+      "PTn = 63.5;         #POtential Transformer\n",
+      "\n",
+      "# Calculations\n",
+      "#R Value of the dial\n",
+      "Rset = (CTp/PTn)*Reff;\n",
+      "Rsetpu = Rset/Vb;\n",
+      "\n",
+      "#X value of the dial\n",
+      "Xset = (CTp/PTn)*Xeff;\n",
+      "Xsetpu = Xset/Vb;\n",
+      "\n",
+      "VRP = 1.0138;         #Assumption Made on the Regulating Point\n",
+      "#Output Voltage of the Regulator\n",
+      "Vreg = VRP+((S/Vln)*((Rset*pf)+(Xset*math.sin(math.radians(math.acos(pf)))))/(CTp*Vb)); \n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Regulating Voltage is %g pu V'%(Vreg)\n",
+      "print 'As per Table 9-6; the primary voltage criteria are met by using the R and X settings'\n",
+      "print 'b The Voltage Profiles are given in the result file attached'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Regulating Voltage is 1.03994 pu V\n",
+        "As per Table 9-6; the primary voltage criteria are met by using the R and X settings\n",
+        "b The Voltage Profiles are given in the result file attached\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6  Page No : 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#From Problems 4 and 5 the co-effcients are obtained\n",
+      "VRRpu = 1.035;\n",
+      "Vreg4 = 1.0337;\n",
+      "Vreg5 = 1.0666;\n",
+      "VRP4 = 1.0337;\n",
+      "VRP5 = 1.0138;\n",
+      "Vmin = 1.010;         #For s =  1.75\n",
+      "\n",
+      "# Calculations\n",
+      "#Steps\n",
+      "Buck4 = (VRRpu-VRP4)/(0.00625);\n",
+      "Buck5 = (VRRpu-VRP5)/(0.00625);\n",
+      "Boost4 = (Vreg4-Vmin)/(0.00625);\n",
+      "Boost5 = (Vreg5-Vmin)/(0.00625);\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Number of steps of buck and number is steps of boost in example 9-4 is %g and %g respectively'%(Buck4,Boost4)\n",
+      "print 'b) The Number of steps of buck and number is steps of boost in example 9-5 is %g and %g respectively'%(Buck5,Boost5)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Number of steps of buck and number is steps of boost in example 9-4 is 0.208 and 3.792 respectively\n",
+        "b) The Number of steps of buck and number is steps of boost in example 9-5 is 3.392 and 9.056 respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8  Page No : 482"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "l = 3.;        #Length of the line\n",
+      "Vlc = 2450.;         #Regulated Voltage\n",
+      "Vcp = 12800.;         #Primary of customer transformer\n",
+      "        #Base Values\n",
+      "Vlbp = 2400.;         #Primary Bus Voltage of Customer's Bus(Low Voltage)\n",
+      "Vlbs = 4160.;        #Secondary Bus Voltage of Customer's Bus\n",
+      "Sb = 5000.;         #Power in kVA\n",
+      "r = 0.3;         #Line Resismath.tance per mile\n",
+      "x = 0.8;         #Line Reacmath.tance per mile\n",
+      "Vhbp = 7390.;        #Primary Bus Voltage of High Voltage Bus\n",
+      "Vhbs = 12800.;        #Secondary Bus Voltage of High Voltage Bus\n",
+      "PTn = 63.5;         #Potential Transformer Turns Ratio\n",
+      "CTp = 250.;        #Current Transformer Turns Ratio\n",
+      "VRP = Vlc/Vlbp;        #Voltage at RP \n",
+      "Vll = Vhbs/1000;         #Line Voltage\n",
+      "VBsec = Vcp/(math.sqrt(3)*PTn);         #Secondary Reading of the Customer Transformer\n",
+      "\n",
+      "# Calculations\n",
+      "VRRset = VRP*VBsec;         #Setting of the voltage-setting dial of VRR\n",
+      "\n",
+      "Zb = (Vll**2)*1000/Sb;         #Applicable Impedance Base\n",
+      "Ztpu = 0.05*1j;        #Transformer Impedance per unit\n",
+      "Zt = Ztpu*Zb;        #Transformer Impedance\n",
+      "\n",
+      "        #Effective Resismath.tances and Reacmath.tances\n",
+      "Reff = (r*l)+Zt.real;\n",
+      "Xeff = (x*l)+Zt.imag;\n",
+      "\n",
+      "Rset = CTp*Reff/PTn;         #X Dial Setting of LDCs\n",
+      "Xset = CTp*Xeff/PTn;         #X Dial Setting of LDCs\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Necessary Setting of the voltage-setting dial of the VRR of each single phase regulator in use is %g V'%(VRRset)\n",
+      "print 'b) R and X dial settings of LDS is %g ohm and %g ohm respectively'%(Rset,Xset)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Necessary Setting of the voltage-setting dial of the VRR of each single phase regulator in use is 118.804 V\n",
+        "b) R and X dial settings of LDS is 3.54331 ohm and 15.8992 ohm respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9  Page No : 484"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Determine the Design Parameters of a Distributed System\n",
+      "#Page 484\n",
+      "\n",
+      "# Variables\n",
+      "VPpu = 1.035;         #Primary Feeder Voltage per unit\n",
+      "TVDpu = 0.0776;         #Total Voltage Drop of Feeder\n",
+      "Vll = 13.2;         #Line Voltage in kV\n",
+      "Vlpuqsw = 1;        #New Voltage at the End of the Feeder due to Qsw at annual peak load\n",
+      "XL = 0.661;         #Inductive Reacmath.tance per mile\n",
+      "Pl = 3400;         #Real Power\n",
+      "Ql = 2100;         #Reactive Power\n",
+      "l = 10.;         #Length of the Feeder in Miles\n",
+      "Lf = 0.4;         #Load Factor\n",
+      "CR = 0.27;         #Total Capacitor Compensation Ratio For the Above Load Factor\n",
+      "QNSW = CR*Ql;         #Required Size of the Nonswitched capacitor bank\n",
+      "s = 2*l/3;         #Loacation of Nonswitched capacitor bank for Optimum Result\n",
+      "VRpu = QNSW*(2*XL*l/3)/(1000*(Vll**2));         #Per Unit Voltage Rise\n",
+      "VDspu = TVDpu*s*(2-(s/l))/l;         #Voltage drop for the uniformaly distributed load\n",
+      "\n",
+      "VSpu = VPpu-VDspu;        #Feeder Voltage at 2l/3 dismath.tance\n",
+      "\n",
+      "nVSpu = VSpu+VRpu;         #New Voltage Rise when there is a fixed capacitor bank\n",
+      "\n",
+      "Vlpu = VPpu-TVDpu;         #When No Capcacitor bank is there, the voltage at the end of the feeder\n",
+      "\n",
+      "nVlpu = Vlpu+VRpu;         #When Capcacitor bank is there, the voltage at the end of the feeder\n",
+      "VRpuqsw = Vlpuqsw-nVlpu;         #Required Voltage Rise\n",
+      "\n",
+      "Q3phisw = 1000*(Vll**2)*VRpuqsw/(XL*l);         #Required Size of the Capacitor Bank\n",
+      "\n",
+      "# Calculations\n",
+      "#Let us assume the 15 single phase standard 50 kVAr Capacitor Units  =  750 kVAr\n",
+      "\n",
+      "SQ3phisw = 750;         #Selected Capacitor Bank\n",
+      "\n",
+      "RVRlpu = VRpuqsw*SQ3phisw/Q3phisw;         #Resulmath.tant Voltage Rises at dismath.tance l\n",
+      "RVRspu = RVRlpu*s/l;         #Resulmath.tant Voltage Rises at dismath.tance s\n",
+      "\n",
+      "#At Peak Load when both the Non-Switched and Switched Capacitor Banks are on\n",
+      "\n",
+      "PVspu = nVSpu+RVRspu;         #Voltage at s\n",
+      "PVlpu = nVlpu+RVRlpu;         #Voltage at l\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The NSW Capacitor Rating is %g kVAr, Which means 2 100kVAr Capacitor Banks per phase'%(QNSW)\n",
+      "print 'b) Voltage Rise per unit is %g pu V'%(VRpu)\n",
+      "print 'i When the No Capacitor Bank is Installed '\n",
+      "print 'Voltage at %g miles is %g pu V'%(s,VSpu)\n",
+      "print 'Voltage at %g miles is %g pu V'%(l,Vlpu)\n",
+      "print 'ii When the Fixed Capacitor Bank is Installed '\n",
+      "print 'Voltage at %g miles is %g pu V'%(s,nVSpu)\n",
+      "print 'Voltage at %g miles is %g pu V'%(l,nVlpu)\n",
+      "print 'c) At Annual Peak Load, The Size of Capacitor Bank Required is %g'%(Q3phisw)\n",
+      "print 'The Voltage Rise at the Annual Load for the Required Capacitor Bank is %g pu V'%(VRpuqsw)\n",
+      "\n",
+      "#Note That The Calculations are highly accurate, Rounding of Terms hasn't be emplyed\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The NSW Capacitor Rating is 567 kVAr, Which means 2 100kVAr Capacitor Banks per phase\n",
+        "b) Voltage Rise per unit is 0.0143399 pu V\n",
+        "i When the No Capacitor Bank is Installed \n",
+        "Voltage at 6.66667 miles is 0.966022 pu V\n",
+        "Voltage at 10 miles is 0.9574 pu V\n",
+        "ii When the Fixed Capacitor Bank is Installed \n",
+        "Voltage at 6.66667 miles is 0.980362 pu V\n",
+        "Voltage at 10 miles is 0.97174 pu V\n",
+        "c) At Annual Peak Load, The Size of Capacitor Bank Required is 744.939\n",
+        "The Voltage Rise at the Annual Load for the Required Capacitor Bank is 0.0282601 pu V\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10  Page No : 488"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "#To Determine the proper 3 phase capacitor bank\n",
+      "#Page 488\n",
+      "\n",
+      "# Variables\n",
+      "V = 12.8;         #Voltage in kV\n",
+      "xl = 0.8;         #Reacmath.tance per unit length\n",
+      "l = 3;         #Dismath.tance of the line\n",
+      "Xl = xl*l;         #Effective Reacmath.tance of the the Line\n",
+      "pf = 0.8;         #Initial Power Factor\n",
+      "pfn = 0.88;         #New Improved Power Factor\n",
+      "Qcu = 150;         #Capacity of each unit available\n",
+      "XT = 1.6384;         #Reacmath.tance of the transformer\n",
+      "\n",
+      "# Calculations\n",
+      "S3phi = 5000*exp(1j*math.pi*math.acos(pf)/180);         #Presently existing Load\n",
+      "\n",
+      "#For New Load Real part of the Load doesn't Change\n",
+      "\n",
+      "QLnew = (S3phi.real)*math.degrees(math.atan(math.acos(pfn)));         #The Required VAr\n",
+      "\n",
+      "S3phinew  = math.sqrt(((S3phi.real)**2)+(QLnew**2));         #New Apparent Power\n",
+      "\n",
+      "Qc = (S3phi.imag)-QLnew;         #Minimum Size of capacitor bank;\n",
+      "\n",
+      "N = math.ceil(Qc/Qcu);         #Number of Units Required\n",
+      "Qcn = N*Qcu;         #Required VAr\n",
+      "\n",
+      "XL = Xl+XT;         #Total Reacmath.tance\n",
+      "\n",
+      "VRpu = Qcn*XL/(1000*(V**2));         #Voltage Rise Per unit\n",
+      "\n",
+      "# Results\n",
+      "print 'The The Voltage Rise found out is %g pu V, which is greater than the voltage rise criterion.Hence %g Capacitor units\\\n",
+      " of %g kVAr must be installed'%(VRpu,N,Qcu)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The The Voltage Rise found out is -3.2425 pu V, which is greater than the voltage rise criterion.Hence -877 Capacitor units of 150 kVAr must be installed\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11  Page No : 493"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Skva = 6.3*(10**3);         #Starting kVA per HP of the Motor\n",
+      "HPmotor = 10.;         #Power Rating\n",
+      "Vll = 7.2*(10**3);         #Line Voltage\n",
+      "I3phi = 1438.;         #Fault Current\n",
+      "\n",
+      "# Calculations\n",
+      "Sstart = Skva*HPmotor;         #Starting kVA\n",
+      "VDIP = 120*Sstart/(I3phi*Vll);         #Voltage Dip\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Voltage dip due to the motor start is %g V'%(VDIP)\n",
+      "print 'b) From the Permissible voltage flicker limit curve%( The Voltage dip of 0.73 Vwith a frequency of\\\n",
+      " 15 times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Voltage dip due to the motor start is 0.730181 V\n",
+        "b) From the Permissible voltage flicker limit curve%( The Voltage dip of 0.73 Vwith a frequency of 15 times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12  Page No : 495"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Skva = 5.6*(10**3);         #Starting kVA per HP of the Motor\n",
+      "HPmotor = 100.;         #Power Rating\n",
+      "Vll = 12.47*(10**3);         #Line Voltage\n",
+      "I3phi = 1765.;         #Fault Current\n",
+      "\n",
+      "# Calculations\n",
+      "Sstart = Skva*HPmotor;         #Starting kVA\n",
+      "VDIP = 69.36*Sstart/(I3phi*Vll);         #Voltage Dip\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Voltage dip due to the motor start is %g V'%(VDIP)\n",
+      "print 'b) From the Permissible voltage flicker limit curve, The Voltage dip of 1.72 Vwith a frequency of three\\\n",
+      " times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Voltage dip due to the motor start is 1.76476 V\n",
+        "b) From the Permissible voltage flicker limit curve, The Voltage dip of 1.72 Vwith a frequency of three times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
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diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/screenshots/loadvstime2.png b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/screenshots/loadvstime2.png
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+++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/screenshots/loadvstime2.png
author	FOSSEE SysAds	2015-12-08 15:04:13 +0600
committer	FOSSEE SysAds	2015-12-08 15:04:13 +0600
commit	534e42c1dee8fefa92f28d4496f273f8c6e4bd94 (patch)
tree	2896915c480bcca6cfbe6432f3fa84e2166f589b /Electric_Power_Distribution_System_Engineering_by_T._Gonen
parent	3ed3fb328a5f4530eec6591d0ca6fe99c69b1013 (diff)
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