{ "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": {} } ] }