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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 09 : Phase Transformations"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.1, Page No 206"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"del_t1 = 0\t\t# temperature difference in degree Celsius\n",
"del_t2 = -5 # temperature difference in degree Celsius\n",
"del_t3 = -40 # temperature difference in degree Celsius\n",
"del_h = 6.02 # enthalpy of fusion in kJ/mol\n",
"T_m = 273.0 # mean temperature\n",
"Gamma = 0.076 # energy of ice water interface in J /m^2\n",
"v = 19.0 # molar volume of ice\n",
"\n",
"#Calculations\n",
"print(\" Part A\")\n",
"print(\" At del_t = %d, there is no supercooling. So there is no critical radius\" %del_t1)\n",
"print(\" Part B\")\n",
"del_f = 16.0/3*math.pi*(Gamma)**3*T_m**2/((del_h*1e3*1e6/v)**2*del_t2**2)\n",
"r = 2*T_m*Gamma/(-del_h*1e3*1e6/v*del_t2)\n",
"print(\" Critical free energy of nucleation is %.1eJ\" %del_f)\n",
"print(\" Critical radius is %d angstrom\" %math.ceil(r*1e10))\n",
"print(\" Part C\")\n",
"temp_r = del_t3/del_t2\n",
"del_f_ = del_f/temp_r**2\n",
"r_ = r/temp_r\n",
"\n",
"#Results\n",
"print(\" Critical free energy of nucleation is %.1eJ\" %del_f_)\n",
"print(\" Critical radius is %d angstrom.\" %math.ceil(r_*1e10))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Part A\n",
" At del_t = 0, there is no supercooling. So there is no critical radius\n",
" Part B\n",
" Critical free energy of nucleation is 2.2e-16J\n",
" Critical radius is 262 angstrom\n",
" Part C\n",
" Critical free energy of nucleation is 3.4e-18J\n",
" Critical radius is 33 angstrom.\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.2, Page No 208"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"T= 300.0 # temperature in kelvin\n",
"R = 8.314 # universal gas constant\n",
"k = 2.303 # conversion factor\n",
"a1 = 1e42\n",
"a2 = 1e6 \t # nucleation rate\n",
"a3 = 1e10 # nucleation rate\n",
"\n",
"#Calculations\n",
"I1 = (math.log(a1,10)-math.log(a2,10))*k*R*T #exponent factor\n",
"I2 = (math.log(a1)-math.log(a3))*k*R*T # exponent factor\n",
"del_f = I1-I2 # difference \n",
"a = 10**(math.log(a3,10)-math.log(a2,10))\n",
"\n",
"#Results\n",
"print(\"A change of %d KJ mol^-1 energy is required to increase nucleation factor from \\n %.0e m^-3 s^-1 to %.0e m^-3 s^-1 \" %(math.ceil(del_f/1e3),a,a3))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"A change of -216 KJ mol^-1 energy is required to increase nucleation factor from \n",
" 1e+04 m^-3 s^-1 to 1e+10 m^-3 s^-1 \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.4, Page No 211"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Gamma_alpha_del = 0.5 # in J m^-2\n",
"Gamma_alpha_beta = 0.5 # in J m^-2\n",
"Gamma_beta_del = 0.01 # in J m^-2\n",
"\n",
"#Calculations\n",
"theta = math.acos((Gamma_alpha_del -Gamma_beta_del)/Gamma_alpha_beta)\n",
"del_f_ratio = 1.0/4*(2-3*math.cos(theta)+(math.cos(theta))**3)\n",
"\n",
"#Results\n",
"print(\"del_f_het is %0.4f th fraction of del_f_homo.\" %del_f_ratio)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"del_f_het is 0.0003 th fraction of del_f_homo.\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.6 Page No 229"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"mu = 45.5e9\n",
"b = 2.55e-10\n",
"n1 = 1e9 \t\t# initial dislocation density\n",
"n2 = 1e13 # final dislocation density\n",
"\n",
"#Calculations\n",
"\n",
"E = 1.0/2*mu*b**2*n2\n",
"del_g = E # as difference between initial and final dislocation energy is four order magnitude\n",
"\n",
"#Results\n",
"print(\"Free energy change during recrystallization is %d J m^-3\" %-del_g)\n",
"\n",
"#Numerical value of answer in book is 14800\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Free energy change during recrystallization is -14793 J m^-3\n"
]
}
],
"prompt_number": 16
}
],
"metadata": {}
}
]
}
|
