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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 7 : Diffusion in Solids"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.1 Page No : 207"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"time required for carburization in 142.8 min\n"
]
}
],
"source": [
"\n",
"import math \n",
"from scipy.special import erfinv\n",
"\n",
"# Variables\n",
"D = 1.28*10**(-11);\t\t\t#diffusion coefficient of carbon in given steel in m2/s\n",
"c_s = 0.9;\t\t\t#Surface concentration of diffusion element in the surface\n",
"c_o = 0.2;\t\t\t#Initial uniform concentration of the element in the solid\n",
"c_x = 0.4;\t\t\t#Concentration of the diffusing element at a distance x from thesurface\n",
"x = 0.5*10**(-3);\t\t\t#depth from the surface in m\n",
"\n",
"# Calculation\n",
"#(c_s-c_x)/(c_s-c_o) = erf(x/(2*(D*t)**(1/2)))\n",
"t = (x/(2*erfinv((c_s-c_x)/(c_s-c_o))*D**(1./2)))**2;\t\t\t#time required for carburization(in sec)\n",
"\n",
"# Results\n",
"print 'time required for carburization in %.1f min'%(t/60)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.2 Page No : 208"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" C1 = 0.0\n",
"time required to get a boron content of 1023 atoms per m3 at a depth of 2 micro meter is = 3845 sec\n"
]
}
],
"source": [
"\t\t\t\n",
"import math \n",
"from scipy.special import erfinv\n",
"\n",
"# Variables\n",
"D = 4*10**(-17);\t\t\t#diffusion coefficient of carbon in given steel in m2/s\n",
"c_s = 3*10**26;\t\t\t#Surface concentration of boron atoms in the surface\n",
"c_1 = 0;\t\t\t#Initial uniform concentration of the element in the solid\n",
"c_x = 10**23;\t\t\t#Concentration of the diffusing element at a distance x from thesurface\n",
"x = 2*10**(-6);\t\t\t#depth from the surface in m\n",
"\n",
"# Calculation and Results\n",
"#(c_s-c_x)/(c_s-c_1) = erf(x/(2*(D*t)**(1/2)))\n",
"a = (erfinv((c_s-c_x)/(c_s-c_1)));\n",
"print ' C1 = ',a\n",
"t = (x**2/(D*4*(2.55)**2));\t\t\t#time required to get a boron content of 1023 atoms per m3 at a depth of 2 micro meter\n",
"print 'time required to get a boron content of 1023 atoms per m3 at a depth of 2 micro meter is = %.0f sec'%t\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.3 Page No : 208"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"activation energy = 2.93e-19 J\n",
"constant of the equation = 2.68e-04 m2/s\n",
"diffusion coefficient at 500°C = 3.27e-16 m2/s\n"
]
}
],
"source": [
"\n",
"import math \n",
"\n",
"# Variables\n",
"t_1 = 736.;\t\t\t#Temperature in °C\n",
"t_2 = 782.;\t\t\t#Temperature in °C\n",
"T_1 = t_1+273;\t\t\t#Temperature in K\n",
"T_2 = t_2+273;\t\t\t#Temperature in K\n",
"D_1 = 2.*10**(-13);\t\t\t#Coefficient of diffusion at T_1 (in m2/s)\n",
"D_2 = 5.*10**(-13);\t\t\t#Coefficient of diffusion at T_2 (in m2/s)\n",
"k = 1.38*10**(-23);\t\t\t#in J/K\n",
"\n",
"# Calculation and Results\n",
"#math.log(d_1) = math.log(d_o)-E/(k*T_1)\n",
"#math.log(d_2) = math.log(d_o)-E/(k*T_2)\n",
"E = (math.log(D_1)-math.log(D_2))/((1/(k*T_1))-(1/(k*T_2)));\t\t\t#\n",
"print 'activation energy = %.2e J'%-E\n",
"D_o = 2.*10**(-13)/math.exp(E/(k*T_1));\n",
"print 'constant of the equation = %.2e m2/s'%D_o\n",
"t_4 = 500.;\t\t\t#Temperature in °C\n",
"T_4 = t_4+273;\t\t\t#Temperature in °K\n",
"D_4 = D_o*math.exp(E/(k*T_4));\t\t\t#diffusion coefficient at 500°C\n",
"print 'diffusion coefficient at 500°C = %.2e m2/s'%D_4\n",
"\n",
"# rounding off error"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.4 Page No : 210"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Time at 500*C that will produce the same diffusion as in 600*C in 110.4 Hours\n"
]
}
],
"source": [
"\t\t\t\n",
"import math \n",
"\n",
"# Variables\n",
"D_500 = 4.8*10**(-14);\t\t\t#Diffusion coefficient for copper in aluminimum at 500*C(in m**2/s)\n",
"D_600 = 5.3*10**(-13);\t\t\t#Diffusion coefficient for copper in aluminimum at 600*C(in m**2/s)\n",
"t_600 = 10;\t\t\t#time of diffussion at 600*C(in Hours)\n",
"\n",
"# Calculation\n",
"#D_500*t_500 = D_600*t_600\n",
"t_500 = D_600*t_600/D_500;\t\t\t#time of diffussion at 500*C\n",
"\n",
"# Results\n",
"print 'Time at 500*C that will produce the same diffusion as in 600*C in %.1f Hours'%t_500\n"
]
}
],
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|
