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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": {}
}
]
}
|
