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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Ch-7, Diffusion in solids"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-7.1 page no-203"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#outward flux of copper is\n",
      "Jx=10**21  #per m**2 per sec\n",
      "#concentration of copper at room temp on one side of aluminium which is 3 mm thickness\n",
      "C1=2*10**13  # per m**3\n",
      "dx=3*10**-3  #m\n",
      "#concentration on the other side of aluminium\n",
      "C2=4*10**6  # per m**3\n",
      "#concentration gradient\n",
      "M=(C1-C2)/dx  #m**4\n",
      "#according to fick's law\n",
      "#Jx=-Dx*dM/dx\n",
      "#so Dx will be\n",
      "Dx=-Jx*1/M  #m**2/s\n",
      "print \"the concentration gradient is %0.2e per m**4and diffusivity of copper is %0.2e m**2/s\"%(M,Dx)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the concentration gradient is 6.67e+15 per m**4and diffusivity of copper is -1.50e+05 m**2/s\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-7.2 page no-209"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp, sqrt\n",
      "from scipy.special import erfinv\n",
      "#given\n",
      "#activation energy and constant D0 are given as\n",
      "#for notation only we are using D01 and D02\n",
      "Q1=83*10**3  #J/mol\n",
      "Q2=157*10**3  #J/mol\n",
      "D01=8*10**-7  #m**2/s\n",
      "D02=700*10**-7  #m**2/s\n",
      "#diffusivity of carbon in alpha-iron at 873 K (600+273) can be obtained by the equation\n",
      "T=873  #K\n",
      "#and \n",
      "#gas constant\n",
      "R=8.314  #J/mol K\n",
      "Dx=D01*exp(-Q1/R/T)  #m**2/s\n",
      "#concentration of carbon is given as 0.75% , therefore concentration of steel will be 0.25 %. thus\n",
      "Ms=0.25  #conc. of steel\n",
      "Mc=0.75  #conc. of carbon\n",
      "#time of carburization is 9 hrs\n",
      "t=9*3600  #sec\n",
      "#so \n",
      "C1=Mc\n",
      "C2=C1-Ms\n",
      "#carbon content on the surface\n",
      "M=0.6\n",
      "#we know that diffusion across a common interface is obtained by:-\n",
      "#M=C1-C2*erf(x/2*sqrt(Dx/t))\n",
      "#solving for X where X=erf(x/2*sqrt(Dx/t))\n",
      "X=((C1-M)/C2)\n",
      "#now for x\n",
      "x=erfinv(X)*2*sqrt(Dx*t)*10**4   #m\n",
      "print \"depth of carburization is %0.2f m\"%(x)  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "depth of carburization is 2.88 m\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-7.3 page no-211"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp\n",
      "from scipy.special import erfinv\n",
      "#given\n",
      "#temp\n",
      "T=950+273  #K\n",
      "QA1=83*10**3   #J/mole\n",
      "QA2=157*10**3  #J/mole\n",
      "R=8.314\n",
      "D01=0.008*10**-4  #m**2/s\n",
      "D02=0.7*10**-4  #m**2/s\n",
      "Ms=0.8  #%\n",
      "Mc=0  #%\n",
      "Mx=0.6  #%\n",
      "t=4*3600  #sec\n",
      "D=1.38*10**-11  #m**2/s \n",
      "#diffusivity at 950 degrees celcius\n",
      "Dx=D01*exp(-QA2/R/T)   #m**2/s\n",
      "#now we use fick's solution which is given by\n",
      "#M(x,t)=C1-C2*erf(x/2/sqrt(Dt))\n",
      "#first boundary condition is \n",
      "#M(x,0)=0.8 for x<0\n",
      "#on solving we get\n",
      "#C1+C2=0.8\n",
      "#second boundary condition is \n",
      "#M(0,t)=0\n",
      "#so we get\n",
      "#C1-C2=0\n",
      "#therefore we get\n",
      "C1=0\n",
      "C2=0.8\n",
      "#so we get x as\n",
      "x=erfinv((C1-Mx)/C2)*2*sqrt(D*t)\n",
      "print \"post machining is to be done upto %0.2e mm\"%x "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "post machining is to be done upto -7.25e-04 mm\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-7.4 page no-213"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from scipy.special import erfinv\n",
      "#given\n",
      "D0=0.002*10**-6  #m**2/s\n",
      "Q=120*10**3  #J/mol\n",
      "T=550+273  #K\n",
      "#the condition for diffusion are\n",
      "#M(x,0)=4% for x>0\n",
      "#M(x,0)=0% for x<0\n",
      "x=(0.25-0.10)*10**-3  #m\n",
      "Mx=0.4  #%\n",
      "R=8.314 #gas constant\n",
      "#diffusion at 550 degrees celcius is\n",
      "Dx=D0*exp(-Q/R/T)  #m**2/s\n",
      "#by using\n",
      "#M(x,t)=C1-C2*erf(x/2/sqrt(Dx*t))\n",
      "#putting boundary conditions in above equation we get\n",
      "C1=2   #%\n",
      "C2=2  #%\n",
      "#solving for t\n",
      "#we get\n",
      "t=(erfinv((C1-Mx)/C2)*2*sqrt(Dx)/x)**-2 #sec\n",
      "print \"the duration for which he material is kept at 550 degrees celcius is %0.2e sec\"%(t)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the duration for which he material is kept at 550 degrees celcius is 1.42e+08 sec\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-7.5 page no-215"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from scipy.special import erfinv\n",
      "#given\n",
      "Mx=10**22  #atoms/m**3\n",
      "M0=10**24  #atoms/m**3\n",
      "t=3*3600  #sec\n",
      "T=1100+273  #K\n",
      "D=7*10**-17   #m**2/s\n",
      "#boundary cinditions are\n",
      "#M(x,0)=0 for x>0 at t=0\n",
      "#M(0,t)=10**24 atoms/m**3\n",
      "#M(x,t)=C1-C2*erf(x/22/sqrt(D*t))\n",
      "#putting boundary conditions in the above equation\n",
      "#we get \n",
      "C1=10**24  #atoms/m**3\n",
      "C2=10**24  #atoms/m**3\n",
      "#solving for x\n",
      "#now for x we have\n",
      "x=erfinv((C1-Mx)/C2)*2*sqrt(D*t)*10**3   #mm\n",
      "print \"depth below the surface at which the concentration is 10**22 atoms/m**3 is %f mm\"%(x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "depth below the surface at which the concentration is 10**22 atoms/m**3 is 0.003167 mm\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-7.6 page no-216"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#temp and pressure of Nitrogen gas\n",
      "T=400  #K\n",
      "P=15 #atm\n",
      "#Nitrogen conc at the inner surface of the tank\n",
      "M=12  #kg/m**3\n",
      "#constant D0 and activation energy\n",
      "D0=5*10**-7  # m**2/s\n",
      "Q=75*10**3  #J/mol\n",
      "#thickness of tank wall\n",
      "x=6*10**-3  #m\n",
      "D=M/x #kg/m**4  (calculation mistake in book)\n",
      "#gas constant\n",
      "R=8.314  #J/mol/K\n",
      "#dufusing flux\n",
      "#Jx=-Dx*m/x kg/m**2/s\n",
      "#Dx can be determined as follows\n",
      "Dx=D0*exp(-Q/R/T)  #m**2/s\n",
      "#so #diffusing flux will be\n",
      "Jx=Dx*D  #kg/m**2/s  (calculation mistake in book)\n",
      "print \"diffusing rate of nitrogen is %0.2e kg/m**2/s\"%(Jx)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "diffusing rate of nitrogen is 1.61e-13 kg/m**2/s\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-7.7 page no-217"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "from __future__ import division\n",
      "#given\n",
      "#ratio of rate of diffusion at 1350 and 110 degree C is 8\n",
      "#so\n",
      "#dM/dt at 1350 degree C(1623K)/dM/dt at 110degree C (1373K)=8\n",
      "#so \n",
      "Ratio=8\n",
      "#we know that\n",
      "#Dx=D0*exp(-Q/R/T)\n",
      "#so Dx (at 1623 K)/Dx (at 1373 K)= exp(-Q/R/T1-Q/R/T2)\n",
      "#where\n",
      "T1=1623  #K\n",
      "T2=1373  #K\n",
      "#gas constant\n",
      "R=8.314  #J/mol/K\n",
      "#so we get\n",
      "#Ratio=exp(-Q/R*(1/T1-1/T2))\n",
      "#so we get Q as\n",
      "Q=-log(Ratio)*R/(1/T1-1/T2)/1000  #KJ/mol\n",
      "print \"the activation energy for sulphur diffusion is %d KJ/mol\"%(Q)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the activation energy for sulphur diffusion is 154 KJ/mol\n"
       ]
      }
     ],
     "prompt_number": 19
    }
   ],
   "metadata": {}
  }
 ]
}