From 47d7279a724246ef7aa0f5359cf417992ed04449 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Wed, 3 Jun 2015 15:27:17 +0530 Subject: add books --- sample_notebooks/MayurSabban/Chapter6.ipynb | 248 ++++++++++++++++++++++++++++ 1 file changed, 248 insertions(+) create mode 100755 sample_notebooks/MayurSabban/Chapter6.ipynb (limited to 'sample_notebooks/MayurSabban') diff --git a/sample_notebooks/MayurSabban/Chapter6.ipynb b/sample_notebooks/MayurSabban/Chapter6.ipynb new file mode 100755 index 00000000..ffc62fd3 --- /dev/null +++ b/sample_notebooks/MayurSabban/Chapter6.ipynb @@ -0,0 +1,248 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 06 : Crystal Imperfection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1, Page No 148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "t1 = 0 #temperature in kelvin\n", + "t2 = 300.0 #temperature in kelvin\n", + "t3 = 900.0 #temperature in kelvin\n", + "R = 8.314 #universal gas constant\n", + "del_hf_al = 68.0 #Enthalpy of formation of aluminium crystal in KJ\n", + "del_hf_ni = 168.0 #Enthalpy of formation of nickel crystal in KJ\n", + "print(\" Example 6.1\")\n", + "\n", + "#Calculations\n", + "print(\"Equilibrium concentration of vacancies of aluminium at %.2f K is 0\" %t1)\n", + "n_N = math.exp(-del_hf_al*1e3/(R*t2))\n", + "print(\" Equilibrium concentration of vacancies of aluminium at %.2f K \" %t2) # answer in book is 1.45e-12\n", + "print(\"is %.2e\" %(n_N))\n", + "n_N = math.exp(-del_hf_al*1e3/(R*t3))\n", + "print(\" Equilibrium concentration of vacancies of aluminium at %.2f K \" %t3) # answer in book is 1.12e-4\n", + "print(\"is %.2e \" %(n_N))\n", + "\n", + "#Results\n", + "print(\"Equilibrium concentration of vacancies of Nickel at %.2f K is 0\" %t1)\n", + "n_N = math.exp(-del_hf_ni*1e3/(R*t2))\n", + "print(\" Equilibrium concentration of vacancies of Nickel at %.2fK \" %t2) # answer in book is 1.45e-12\n", + "print(\"is %.2e\" %(n_N))\n", + "n_N = math.exp(-del_hf_ni*1e3/(R*t3))\n", + "print(\" Equilibrium concentration of vacancies of Nickel at %.2f K \" %t3) # answer in book is 1.78e-10\n", + "print(\"is %.2e \" %(n_N))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Example 6.1\n", + "Equilibrium concentration of vacancies of aluminium at 0.00 K is 0\n", + " Equilibrium concentration of vacancies of aluminium at 300.00 K \n", + "is 1.44e-12\n", + " Equilibrium concentration of vacancies of aluminium at 900.00 K \n", + "is 1.13e-04 \n", + "Equilibrium concentration of vacancies of Nickel at 0.00 K is 0\n", + " Equilibrium concentration of vacancies of Nickel at 300.00K \n", + "is 5.59e-30\n", + " Equilibrium concentration of vacancies of Nickel at 900.00 K \n", + "is 1.77e-10 \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2, Page No 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "a = 2.87 #lattice parameter in angstrom\n", + "b= 2.49 #magnitude of burgers vector in angstrom\n", + "G = 80.2 #shear modulus in GN\n", + "\n", + "#Calculations\n", + "E = G*1e9*(b*1e-10)**2*1.0/2 \n", + "\n", + "#Results\n", + "print(\"Line energy of dislocation is %.2e J m^-1\" %E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Line energy of dislocation is 2.49e-09 J m^-1\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, Page No 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "a = 1.0e10 # total number of edge dislocation \n", + "N = 6.023e23 \t# Avogadro number\n", + "R = 8.314 \t# Universal gas constant\n", + "t1 = 0 \t# initial temperature in K\n", + "t2 = 1000.0\t # Final temperature in K\n", + "del_hf = 100.0 \t # Enthalpy of vacancy formation in KJ\n", + "d = 2.0 # length of one step in angstrom\n", + "\n", + "#Calculations\n", + "v = 5.5/10**6 # volume of one mole crystal\n", + "n = N*math.exp(-(del_hf*1e03)/(R*(t2-t1)))/v\n", + "k = 1.0/(d*1e-10 ) # atoms required for 1 m climb\n", + "b = n/(k*a)# average amount of climb\n", + "c = b*d*1e-10 \n", + "\n", + "#Results\n", + "print(\"Average down climb of crystal is %.2e m\" %c)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average down climb of crystal is 2.62e-06 m\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 Page No 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "\n", + "E = 56.4 #bond energy in KJ\n", + "N_a = 6.023e23 #Avogadro\u2019s number\n", + "n = 12.0 #number of bonds\n", + "m = 3.0 #number of broken bonds \n", + "N = 1.77e19 #number of atoms in copper crystal of type {111} per m^2\n", + "\n", + "#Calculations\n", + "b_e = 1.0/2*E*1e3*n/N_a #bond energy per atom\n", + "e_b = b_e*m/n #energy of broken bond at surface\n", + "s_e = e_b*N #surface enthalpy of copper\n", + "\n", + "#Results\n", + "print(\"Surface enthalpy of copper is %0.2f J m^-2\" %s_e)\n", + "print(\"Surface enthalpy of copper is %0.2f erg cm^-2\" %(s_e*1e3))\n", + "# Answer in book is 2490 erg cm^-2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface enthalpy of copper is 2.49 J m^-2\n", + "Surface enthalpy of copper is 2486.17 erg cm^-2\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, Page No 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Gamma_gb = 1.0 #let, energy of grain boundary\n", + "\n", + "#Calculations\n", + "Gamma_s = 3.0*Gamma_gb #energy of free surface\n", + "theta = 2*math.degrees(math.acos(1.0/2*Gamma_gb/Gamma_s))\n", + "\n", + "#Results\n", + "print(\"Angle at the bottom of groove of a boundary is %0.2f degrees.\" %math.ceil(theta))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle at the bottom of groove of a boundary is 161.00 degrees.\n" + ] + } + ], + "prompt_number": 28 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit