{ "metadata": { "name": "", "signature": "sha256:c24c5f13f06ad2e35494c2ba2e22ba1351ce2de7d258eb50313205ed297dec4a" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4: Diffusion In Solids" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1: Page 89" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Illustration 4.1\n", "# Page: 89\n", "\n", "import math\n", "print'Illustration 4.1 - Page: 89\\n\\n'\n", " \n", "# solution\n", "\n", "#***Data****#\n", "P = 2;# [atm]\n", "a1 = 0.025;# [m]\n", "a2 = 0.050;# [m]\n", "solub = 0.053*P;# [cubic m H2 (STP)/(cubic m rubber)]\n", "Ca1 = solub/22.41;# inner surface of the pipe\n", "Ca2 = 0;# resistance to difusion of H2 away from the surface is negligible.\n", "Da = 1.8*10**(-10);# [square m/s]\n", "l = 1;# [m]\n", "#********#\n", "\n", "z = (a2-a1)/2;# [m]\n", "# Using Eqn. 4.4\n", "Sav = (2*(math.pi)*l*(a2-a1))/(2*math.log(a2/a1));# [square m]\n", "# Using Eqn. 4.3\n", "w = (Da*Sav*(Ca1-Ca2))/z;# [kmol H2/s for 1m length]\n", "w = w*2.02*10**3*3600;# [g H2/m.h]\n", "print'The rate of loss of H2 by diffusion per m of pipe length:',round(w,6),' g H2/m.h'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Illustration 4.1 - Page: 89\n", "\n", "\n", "The rate of loss of H2 by diffusion per m of pipe length: 5.6e-05 g H2/m.h\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.2: Page 92" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Illustration 4.2\n", "# Page: 92\n", "\n", "print'Illustration 4.2 - Page: 92\\n\\n'\n", "print'Illustration 4.2 (a)\\n\\n'\n", "\n", "# solution (a)\n", "\n", "# Given\n", "a = 3.0/2;# [cm]\n", "thetha = 68*3600;# [s]\n", "# Ca can e calculated in terms of g/100 cubic cm\n", "Cao = 5.0;# [g/100 cubic cm]\n", "Ca_thetha = 3.0;# [g/100 cubic cm]\n", "Ca_Inf = 0.0;# [g/100 cubic cm]\n", "#**********#\n", "\n", "E = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);\n", "# E = 0.6;\n", "# From Fig. 4.2 (Pg 91): For diffusion from only one exposed surface D*thetha/(4*a^2) = 0.128\n", "D = 0.128*4*(a**2)/thetha;# [square cm/s]\n", "D = D*10**(-4);# [square m/s]\n", "print'Diffusivity of urea in gel from only one exposed durface:',round(D,12),'square m/s\\n\\n'\n", "\n", "print'Illustration 4.2 (b)\\n\\n'\n", "\n", "# Solution (b)\n", "\n", "#****Data****#\n", "# Ca can e calculated in terms of g/100 cubic cm\n", "Cao = 5.0;# [g/100 cubic cm]\n", "Ca_thetha = 1.0;# [g/100 cubic cm]\n", "Ca_Inf = 0.0;# [g/100 cubic cm]\n", "#*********#\n", "\n", "E = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);\n", "# E = 0.2;\n", "# From Fig. 4.2 (Pg 91): For diffusion from only one exposed surface D*thetha/(4*a**2) = 0.568\n", "D = 4.70*10**(-6);# From Illusration 4.2(a) [square cm/s]\n", "thetha = 0.568*4*a**2/D;# [s]\n", "thetha = thetha/3600.0;# [h]\n", "print'The time taken for the avg. conc. to fall to 1g/100 cubic cm is:',round(thetha),' hours'\n", "\n", "print'Illustration 4.2 (c)\\n\\n'\n", "\n", "# solution (c)\n", "\n", "#****Data*****#\n", "Cao = 5.0;# [g/100 cubic cm]\n", "Ca_thetha = 1.0;# [g/100 cubic cm]\n", "Ca_Inf = 0.0;# [g/100 cubic cm]\n", "#*******#\n", "\n", "E = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);\n", "# E = 0.2;\n", "# From Fig. 4.2: For diffusion from two opposite exposed surface D*thetha/(a**2) = 0.568\n", "D = 4.70*10**(-6);# From Illusration 4.2(a) [square cm/s]\n", "thetha = 0.568*(a**2)/D;# [s]\n", "thetha = thetha/3600.0;# [h]\n", "print'The time taken for the avg. conc. to fall to 1g/100 cubic cm when two faces opposed is:',int(thetha),' hours'\n", "# the solution in the textbook is wrong due to approximation\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Illustration 4.2 - Page: 92\n", "\n", "\n", "Illustration 4.2 (a)\n", "\n", "\n", "Diffusivity of urea in gel from only one exposed durface: 4.71e-10 square m/s\n", "\n", "\n", "Illustration 4.2 (b)\n", "\n", "\n", "The time taken for the avg. conc. to fall to 1g/100 cubic cm is: 302.0 hours\n", "Illustration 4.2 (c)\n", "\n", "\n", "The time taken for the avg. conc. to fall to 1g/100 cubic cm when two faces opposed is: 75 hours\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.3: Page 94" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Illustration 4.3\n", "# Page: 94\n", "\n", "print'Illustration 4.3 - Page: 94\\n\\n'\n", "\n", "# solution \n", "\n", "#****Data****#\n", "z = 0.1;# [cm]\n", "pa1 = 1;# [cmHg]\n", "pa2 = 0;# [cmHg]\n", "Da = 1.1*10**(-10)*10**4;# [square cm/s]\n", "#***********#\n", "\n", "# Solubility coeffecient in terms of Hg\n", "Sa = 0.90/76;# [cubic cm gas (STP)/cubic cm.cmHg]\n", "# Using Eqn. 4.15\n", "Va = (Da*Sa*(pa1-pa2))/z;# [cubic cm(STP)/square cm.s]\n", "# Using Eqn. 4.16\n", "P = Da*Sa;# [cubic cm gas (STP)/square cm.s.(cmHg/cm)]\n", "print'The rate of diffusion of CO is:',round(Va,8),'cubic cm(STP)/square cm.s'\n", "print'The permeability of the membrane is',round(P,9),'cubic cm gas (STP)/square cm.s.(cmHg/cm)'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Illustration 4.3 - Page: 94\n", "\n", "\n", "The rate of diffusion of CO is: 1.3e-07 cubic cm(STP)/square cm.s\n", "The permeability of the membrane is 1.3e-08 cubic cm gas (STP)/square cm.s.(cmHg/cm)\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.4: Page 96" ] }, { "cell_type": "code", "collapsed": false, "input": [ " \n", "\n", "# Illustration 4.4\n", "# Page: 96\n", "\n", "print'Illustration 4.4 - Page: 96\\n\\n'\n", "\n", "# solution\n", "\n", "#****Data****#\n", "a = 0.005;# [m]\n", "# For the KCl diffusion\n", "Dab1 = 1.84*10**(-9);# [square m/s]\n", "thetha = 4.75*3600;# [s]\n", "Ca_Inf = 0;\n", "# For K2CrO4 diffusion\n", "Cao = 0.28;# [g/cubic cm]\n", "Ca_Inf = 0.002;# [g/cubic cm]\n", "Dab2 = 1.14*10**(-9);# [square m/s]\n", "#*******#\n", "\n", "E = 0.1;# For 90% removal of KCl\n", "# From Fig. 4.2 (Pg 91): Deff*thetha/a^2 = 0.18\n", "Deff = 0.18*a**2/thetha;# [square m/s]\n", "Dab_by_Deff = Dab1/Deff;\n", "Ca_thetha = 0.1*0.28;# [g/cubic cm]\n", "Es = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);\n", "# From Fig. 4.2 (Pg 91): Deff*thetha/a^2 = 0.30\n", "Deff = Dab2/Dab_by_Deff;# [square m/s]\n", "thetha = 0.3*a**2/Deff;# [s]\n", "thetha = thetha/3600;# [h]\n", "print'The time reqd. is:',round(thetha,3),'hours'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Illustration 4.4 - Page: 96\n", "\n", "\n", "The time reqd. is: 12.778 hours\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.5: Page 98" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Illustration 4.5\n", "# Page: 98\n", "import math \n", "\n", "print'Illustration 4.5 - Page: 98\\n\\n'\n", "print'Illustration 4.5 (a)\\n\\n'\n", "\n", "# solution (a)\n", "\n", "#****Data****#\n", "# a = H2 b = N2\n", "Dab_eff = 5.3*10**(-6);# [square m/s]\n", "Dkb_eff = 1.17*10**(-5);# [square m/s]\n", "Dab = 7.63*10**(-5);# [square m/s]\n", "#*******#\n", "\n", "R = 8314;#[Nm/kmol]\n", "Mb = 2.02;# [kg/kmol]\n", "T = 293;# [K]\n", "Dtrue_by_Deff = Dab/Dab_eff;\n", "# Since the ratio is strictly a matter of the geometry of the solid.\n", "Dkb = Dkb_eff*Dtrue_by_Deff;# [square m/s]\n", "# From Eqn. 4.20\n", "d = 3*Dkb*((math.pi*Mb)/(8*R*T))**0.5;# [m]\n", "print'The equivalent pore diameter is: ',round(d,9),' m\\n\\n'\n", "\n", "print'Illustration 4.5 (b)\\n\\n'\n", "\n", "# Solution (b)\n", "\n", "#****Data*****#\n", "# a = O2 b = N2 c = H2\n", "Ya1 = 0.8;\n", "Ya2 = 0.2;\n", "Pt = 10133;# [N/square m]\n", "z = 0.002;# [m]\n", "T = 293;# [K]\n", "#*******#\n", "\n", "# From Table 2.1 (Pg 31):\n", "Dab = 1.81*10**(-5);# [square m/s] at STP\n", "Dkc = 1.684*10**(-4);# [square m/s] From Illustration 4.5(a)\n", "Mc = 2.02;# [kg/kmol]\n", "Ma = 32;# [kg/kmol]\n", "Mb = 28.02;# [kg/kmol]\n", "Dab = Dab*(1/0.1)*((293/273)**1.5);# [square m/s] at 0.1 atm & 20 C\n", "DabEff = Dab/14.4;# [square m/s] From Illustration 4.5(a)\n", "Dka = Dkc*((Mc/Ma)**0.5);# [square m/s]\n", "DkaEff = Dka/14.4;# [square m/s]\n", "Nb_by_Na = -(Ma/Mb)**0.5;\n", "# Na/(Na+Nb) = 1.0/(1+(Nb/Na))\n", "Na_by_NaSumNb = 1.0/(1+(Nb_by_Na));\n", "DabEff_by_DkaEff = DabEff/DkaEff;\n", "# By Eqn. 4.23\n", "Na = (Na_by_NaSumNb)*(DabEff*Pt/(R*T*z))*log((((Na_by_NaSumNb)*(1+DabEff_by_DkaEff))-Ya2)/(((Na_by_NaSumNb)*(1+DabEff_by_DkaEff))-Ya1));# [kmol/square m.s]\n", "Nb = Na*(Nb_by_Na);# [kmol/square m.s]\n", "print\"Diffusion flux of O2 is \",round(Na,8),\" kmol/square m.s\\n\"\n", "print\"Diffusion flux of N2 is \",round(Nb,8),\" kmol/square m.s\\n\"\n", "#the answer in textbook is slightly different due to approximation while here calculation is precise" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Illustration 4.5 - Page: 98\n", "\n", "\n", "Illustration 4.5 (a)\n", "\n", "\n", "The equivalent pore diameter is: 2.88e-07 m\n", "\n", "\n", "Illustration 4.5 (b)\n", "\n", "\n", "Diffusion flux of O2 is 2.95e-06 kmol/square m.s\n", "\n", "Diffusion flux of N2 is -3.16e-06 kmol/square m.s\n", "\n" ] } ], "prompt_number": 33 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.6: Page 100" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Illustration 4.6\n", "# Page: 100\n", "\n", "import math\n", "print'Illustration 4.6 - Page: 100\\n\\n'\n", "\n", "# solution\n", "\n", "#***Data***#\n", "# a = N2\n", "# For N2 at 300K\n", "viscosity1 = 1.8*10**(-5);# [kg/m.s]\n", "Pt1 = 10133.0;# [N/square m.sec]\n", "T = 300;# [K]\n", "z = 0.0254;# [m]\n", "T2 = 393.0;# [K]\n", "#***********#\n", "\n", "Ma = 28.02;# [kg/kmol]\n", "R = 8314.0;# [J/K.kgmol]\n", "#From Eqn 4.22\n", "Lambda = (3.2*viscosity1/Pt1)*(R*T/(2*(math.pi)*Ma))**0.5;\n", "d = 10**(-4);# [m]\n", "d_by_lambda = d/Lambda;\n", "# Kundsen flow will not occur\n", "# N2 flow corresponding to 9 cubic ft/square ft.min at 300K & 1 std atm = 0.0457 cubic m/square m.min\n", "Na1 = 0.0457*(273.0/T)*(1/22.41);# [kmol/square m.s]\n", "Pt1_diff_Pt2 = 2*3386/13.6;# [N/square m]\n", "Ptav = Pt1+(Pt1_diff_Pt2/2.0);# [N/square m]\n", "# From Eqn. 4.26\n", "k1 = Na1*R*T*z/(Ptav*(Pt1_diff_Pt2));# [m**4/N.s]\n", "\n", "#For N2 at 393K\n", "viscosity2 = 2.2*10**(-5);# [kg/m.s]\n", "k2 = (k1*viscosity1)/(viscosity2);# [m^4/N.s]\n", "# From Eqn 4.26\n", "Na = (k2*Ptav*Pt1_diff_Pt2)/(R*T2*z);# [kmol/square m.s]\n", "print\"Flow rate to be expected is\",round(Na,6),\" kmol/square m.s\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Illustration 4.6 - Page: 100\n", "\n", "\n", "Flow rate to be expected is 0.001159 kmol/square m.s\n" ] } ], "prompt_number": 44 } ], "metadata": {} } ] }