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+{

+ "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": {}

+  }

+ ]

+}
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