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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5 : Transport with a net convective flux"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.9 - Page No :166\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"# given\n",
"v = 1.; \t\t\t #[cm/sec] - volume velocity or bulk velocity\n",
"vol = 1.; \t\t\t #[cm**3] - volume\n",
"na = 2.; \t\t\t # moles of a\n",
"nb = 3.; \t\t\t # moles of b\n",
"nc = 4.; \t\t\t # moles of c\n",
"mma = 2.; \t\t\t #molecular weight of a\n",
"mmb = 3.; \t\t\t #molecular weight of b\n",
"mmc = 4.; \t\t\t #molecular weight of c\n",
"ma = na*mma; \t\t\t #[g] weight of a\n",
"mb = nb*mmb; \t\t\t #[g] weight of b\n",
"mc = nc*mmc; \t\t\t #[g] weight of c\n",
"NabyA = 2.+2; \t\t\t #[mol/cm**2*s] - molar flux = diffusing flux +convected flux\n",
"NbbyA = -1.+3; \t\t\t #[mol/cm**2*s] - molar flux = diffusing flux +convected flux\n",
"NcbyA = 0.+4; \t\t\t #[mol/cm**2*s] - molar flux = diffusing flux +convected flux\n",
"NtbyA = NabyA+NbbyA+NcbyA; \t\t\t #[mol/cm**2*s] - total molar flux\n",
"\n",
"# Calculations\n",
"# on a mass basis,these corresponds to\n",
"nabyA = 4.+4; \t\t\t #[g/cm**2*s]; - mass flux = diffusing flux +convected flux\n",
"nbbyA = -3.+9; \t\t\t #[g/cm**2*s]; - mass flux = diffusing flux +convected flux\n",
"ncbyA = 0.+16; \t\t\t #[g/cm**2*s]; - mass flux = diffusing flux +convected flux\n",
"ntbyA = nabyA+nbbyA+ncbyA; \t\t\t #[g/cm**2*s] - total mass flux\n",
"\n",
"# concentrations are expressed in molar basis\n",
"CA = na/vol; \t\t\t #[mol/cm**3]\n",
"CB = nb/vol; \t\t\t #[mol/cm**3]\n",
"CC = nc/vol; \t\t\t #[mol/cm**3]\n",
"CT = CA+CB+CC; \t\t\t #[mol/cm**3] - total concentration\n",
"\n",
"# densities are on a mass basis\n",
"pa = ma/vol; \t\t\t #[g/cm**3]\n",
"pb = mb/vol; \t\t\t #[g/cm**3]\n",
"pc = mc/vol; \t\t\t #[g/cm**3]\n",
"pt = pa+pb+pc; \t\t\t #[g/cm**3]\n",
"Ua = NabyA/CA; \t\t\t #[cm/sec];\n",
"Ub = NbbyA/CB; \t\t\t #[cm/sec];\n",
"Uc = NcbyA/CC; \t\t\t #[cm/sec];\n",
"# the same result will be obtained from dividing mass flux by density\n",
"Uz = (pa*Ua+pb*Ub+pc*Uc)/(pa+pb+pc);\n",
"\n",
"# Results\n",
"print \" Uz = %.3f cm/sec\"%(Uz);\n",
"Uzstar = (NtbyA/CT);\n",
"print \" Uz* = %.2f cm/sec\"%(Uzstar);\n",
"print \" For this Example both Uz and Uz* are slightly greater than the volume \\\n",
" velocity of 1cm/sec, because there is a net molar and \\n mass diffusion in the positive direction.\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Uz = 1.034 cm/sec\n",
" Uz* = 1.11 cm/sec\n",
" For this Example both Uz and Uz* are slightly greater than the volume velocity of 1cm/sec, because there is a net molar and \n",
" mass diffusion in the positive direction.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.10 - Page No :171\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"# given (from Example 5.9)\n",
"na = 2.; \t\t\t # moles of a\n",
"nb = 3.; \t\t\t # moles of b\n",
"nc = 4.; \t\t\t # moles of c\n",
"mma = 2.; \t\t\t #molecular weight of a\n",
"mmb = 3.; \t\t\t #molecular weight of b\n",
"mmc = 4.; \t\t\t #molecular weight of c\n",
"ma = na*mma; \t\t\t #[g] weight of a\n",
"mb = nb*mmb; \t\t\t #[g] weight of b\n",
"mc = nc*mmc; \t\t\t #[g] weight of c\n",
"NabyA = 2.+2; \t\t\t #[mol/cm**2*s] - molar flux = diffumath.sing flux +convected flux\n",
"NbbyA = -1.+3; \t\t\t #[mol/cm**2*s] - molar flux = diffusing flux +convected flux\n",
"NcbyA = 0.+4; \t\t\t #[mol/cm**2*s] - molar flux = diffusing flux +convected flux\n",
"NtbyA = NabyA+NbbyA+NcbyA; \t\t\t #[mol/cm**2*s] - total molar flux\n",
"vol= 1.\n",
"# Calculations\n",
"# on a mass basis,these corresponds to\n",
"nabyA = 4+4; \t\t\t #[g/cm**2*s]; - mass flux = diffusing flux +convected flux\n",
"nbbyA = -3+9; \t\t\t #[g/cm**2*s]; - mass flux = diffusing flux +convected flux\n",
"ncbyA = 0+16; \t\t\t #[g/cm**2*s]; - mass flux = diffusing flux +convected flux\n",
"\n",
"# concentrations are expressed in molar basis\n",
"CA = na/vol; \t\t\t #[mol/cm**3]\n",
"CB = nb/vol; \t\t\t #[mol/cm**3]\n",
"CC = nc/vol; \t\t\t #[mol/cm**3]\n",
"CT = CA+CB+CC; \t\t #[mol/cm**3] - total concentration\n",
"\n",
"# densities are on a mass basis\n",
"pa = ma/vol; \t\t\t #[g/cm**3]\n",
"pb = mb/vol; \t\t\t #[g/cm**3]\n",
"pc = mc/vol; \t\t\t #[g/cm**3]\n",
"Ua = NabyA/CA; \t\t\t #[cm/sec];\n",
"Ub = NbbyA/CB; \t\t\t #[cm/sec];\n",
"Uc = NcbyA/CC; \t\t\t #[cm/sec];\n",
"U = (pa*Ua+pb*Ub+pc*Uc)/(pa+pb+pc);\n",
"Ustar = (NtbyA/CT);\n",
"\n",
"# the fluxes relative to mass average velocities are found as follows\n",
"JabyA = CA*(Ua-U); \t\t\t #[mol/cm**2*sec]\n",
"JbbyA = CB*(Ub-U); \t\t\t #[mol/cm**2*sec]\n",
"JcbyA = CC*(Uc-U); \t\t\t #[mol/cm**2*sec]\n",
"\n",
"# Results\n",
"print \" fluxes relative to mass average velocities are-\";\n",
"print \" Ja/A = %.4f mol/cm**2*sec\"%(JabyA);\n",
"print \" Jb/A = %.4f mol/cm**2*sec\"%(JbbyA);\n",
"print \" Jc/A = %.4f mol/cm**2*sec\"%(JcbyA);\n",
"jabyA = pa*(Ua-U); \t\t\t #[g/cm**2*sec]\n",
"jbbyA = pb*(Ub-U); \t\t\t #[g/cm**2*sec]\n",
"jcbyA = pc*(Uc-U); \t\t\t #[g/cm**2*sec]\n",
"print \" ja/A = %.4f g/cm**2*sec\"%(jabyA);\n",
"print \" jb/A = %.4f g/cm**2*sec\"%(jbbyA);\n",
"print \" jc/A = %.4f g/cm**2*sec\"%(jcbyA);\n",
"\n",
"# the fluxes relative to molar average velocity are found as follows\n",
"JastarbyA = CA*(Ua-Ustar); \t\t\t #[mol/cm**2*sec]\n",
"JbstarbyA = CB*(Ub-Ustar); \t\t\t #[mol/cm**2*sec]\n",
"JcstarbyA = CC*(Uc-Ustar); \t\t\t #[mol/cm**2*sec]\n",
"print \" fluxes relative to molar average velocities are-\";\n",
"print \" Ja*/A = %.4f mol/cm**2*sec\"%(JastarbyA);\n",
"print \" Jb*/A = %.4f mol/cm**2*sec\"%(JbstarbyA);\n",
"print \" Jc*/A = %.4f mol/cm**2*sec\"%(JcstarbyA);\n",
"jastarbyA = pa*(Ua-Ustar); \t\t\t #[g/cm**2*sec]\n",
"jbstarbyA = pb*(Ub-Ustar); \t\t\t #[g/cm**2*sec]\n",
"jcstarbyA = pc*(Uc-Ustar); \t\t\t #[g/cm**2*sec]\n",
"print \" ja*/A = %.4f g/cm**2*sec\"%(jastarbyA);\n",
"print \" jb*/A = %.4f g/cm**2*sec\"%(jbstarbyA);\n",
"print \" jc*/A = %.4f g/cm**2*sec\"%(jcstarbyA);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" fluxes relative to mass average velocities are-\n",
" Ja/A = 1.9310 mol/cm**2*sec\n",
" Jb/A = -1.1034 mol/cm**2*sec\n",
" Jc/A = -0.1379 mol/cm**2*sec\n",
" ja/A = 3.8621 g/cm**2*sec\n",
" jb/A = -3.3103 g/cm**2*sec\n",
" jc/A = -0.5517 g/cm**2*sec\n",
" fluxes relative to molar average velocities are-\n",
" Ja*/A = 1.7778 mol/cm**2*sec\n",
" Jb*/A = -1.3333 mol/cm**2*sec\n",
" Jc*/A = -0.4444 mol/cm**2*sec\n",
" ja*/A = 3.5556 g/cm**2*sec\n",
" jb*/A = -4.0000 g/cm**2*sec\n",
" jc*/A = -1.7778 g/cm**2*sec\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.11 - Page No :176\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"# Variables\n",
"# given\n",
"T = 0+273.15; \t\t\t #[K] - temperature in Kelvins\n",
"pa2 = 1.5; \t\t\t #[atm] - partial presuure of a at point2\n",
"pa1 = 0.5; \t\t\t #[atm] - partial pressure of a at point 1\n",
"z2 = 20.; \t\t\t #[cm] - position of point 2 from reference point\n",
"z1 = 0.; \t\t\t #[cm] - position of point1 from reference point\n",
"p = 2.; \t\t\t #[atm] - total pressure\n",
"d = 1.; \t\t\t #[cm] - diameter\n",
"D = 0.275; \t\t #[cm**2/sec] - diffusion coefficient\n",
"A = (math.pi*((d)**2))/4.;\n",
"R = 0.082057; \t\t\t #[atm*m**3*kmol**-1*K**-1] - gas constant\n",
"\n",
"# Calculations\n",
"# (a) using the formula Na/A = -(D/(R*T))*((pa2-pa1)/(z2-z1))\n",
"Na = (-(D/(R*T))*((pa2-pa1)/(z2-z1)))*(A)/(10.**6);\n",
"print \" Na = %.2e kmol/sec \\n The negative sign indicates diffusion from point 2 to point 1\"%(Na);\n",
"pb2 = p-pa2;\n",
"pb1 = p-pa1;\n",
"\n",
"# (b) using the formula Na/A = ((D*p)/(R*T*(z2-z1)))*ln(pb2/pb1)\n",
"Na = (((D*p)/(R*T*(z2-z1)))*math.log(pb2/pb1))*(A)/(10**6);\n",
"\n",
"# Results\n",
"print \" Na = %.2e kmol/sec\"%(Na);\n",
"print \" The induced velocity increases the net transport of A by the ratio of 10.6*10**-10 \\\n",
"to 4.82*10**-10 or 2.2 times.This increse is equivalent to 120 percent\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Na = -4.82e-10 kmol/sec \n",
" The negative sign indicates diffusion from point 2 to point 1\n",
" Na = -1.06e-09 kmol/sec\n",
" The induced velocity increases the net transport of A by the ratio of 10.6*10**-10 to 4.82*10**-10 or 2.2 times.This increse is equivalent to 120 percent\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.12 - Page No :178\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"# given\n",
"T = 0+273.15; \t\t\t #[K] - temperature in Kelvins\n",
"pa2 = 1.5; \t\t\t #[atm] - partial presuure of a at point2\n",
"pa1 = 0.5; \t\t\t #[atm] - partial pressure of a at point 1\n",
"z2 = 20.; \t\t\t #[cm] - position of point 2 from reference point\n",
"z1 = 0.; \t\t\t #[cm] - position of point1 from reference point\n",
"p = 2.; \t\t\t #[atm] - total pressure\n",
"d = 1.; \t\t\t #[cm] - diameter\n",
"D = 0.275; \t\t\t #[cm**2/sec] - diffusion coefficient\n",
"\n",
"# Calculations\n",
"A = (math.pi*((d)**2.))/4;\n",
"R = 0.082057; \t\t\t #[atm*m**3*kmol**-1*K**-1] - gas consmath.tant\n",
"k = 0.75;\n",
"\n",
"# umath.sing the formula (Na/A) = -(D/(R*T*(z2-z1)))*ln((1-(pa2/p)*(1-k))/(1-(pa1/p)*(1-k)))\n",
"NabyA = -(D/(R*T*(z2-z1)))*(2*0.7854)*math.log((1-(pa2/p)*(1-k))/(1-(pa1/p)*(1-k)))/(10**6);\n",
"\n",
"# Results\n",
"print \" Na/A = %.2e kmol/sec\"%(NabyA);\n",
"print \" Note that this answer is larger than the rate for equimolar counter diffusion \\\n",
"but smaller tahn the rate for diffusion through a stagnant film. \\nSometimes the\\\n",
" rate for diffusin through a stagnant film can be considered as an upper bound\\\n",
" if k ties between zero and one\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Na/A = 1.38e-10 kmol/sec\n",
" Note that this answer is larger than the rate for equimolar counter diffusion but smaller tahn the rate for diffusion through a stagnant film. \n",
"Sometimes the rate for diffusin through a stagnant film can be considered as an upper bound if k ties between zero and one\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.13 - Page No :184\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"# given\n",
"l = 4.; \t\t\t #[m] - length of the tube\n",
"id_ = 1.6*10**-3; \t\t\t #[m] - insid_e diameter\n",
"Nkn = 10.; \t\t \t # - knudsen no.\n",
"Ma = 92.; \t\t\t # - molecular weight of gas\n",
"mu = 6.5*10**-4; \t\t\t #[kg/m*sec] - vismath.cosity\n",
"T = 300.; \t \t\t #[K] - temperature\n",
"R = 8314.; \t \t\t #[kPa*m**3*kmol**-1*K**-1] - gas consmath.tant\n",
"lambdaA = Nkn*id_; \t\t\t #[m] mean free path\n",
"\n",
"# Calculations\n",
"# for calculating pressure umath.sing the formula lamdaA = 32*(mu/p)*((R*T)/(2*pi*Ma))**(1/2)\n",
"p = 32*(mu/lambdaA)*((R*T)/(2*math.pi*Ma))**(1/2.);\n",
"patm = p/(1.01325*10**5);\n",
"\n",
"# Results\n",
"print \" p = %.2f kg/m*sec**2 = %.2f Pa = %.2e atm\"%(p,p,patm);\n",
"print \" The value of 10 for the knudsen number is on the border \\\n",
" between Knudsen diffusion and transition flow\";\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" p = 85.39 kg/m*sec**2 = 85.39 Pa = 8.43e-04 atm\n",
" The value of 10 for the knudsen number is on the border between Knudsen diffusion and transition flow\n"
]
}
],
"prompt_number": 11
}
],
"metadata": {}
}
]
}
|
