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"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": {}
}
]
}