{ "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": {} } ] }