{ "metadata": { "name": "", "signature": "sha256:c20c5ffc14398659aab9089e6b803cb3218676cd974382b644456d53b86a097c" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Diffusion Mass Transfer" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.1 Page 884" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "import math\n", "\n", "T = 293. \t \t\t\t\t;#[K] Temperature\n", "Ma = 2 \t\t\t\t\t;#[kg/kmol] Molecular Mass\n", "#Table A.8 Hydrogen-Air Properties at 298 K\n", "Dab1 = .41*math.pow(10,-4); #[m^2/s] diffusion coefficient\n", "#Table A.8 Hydrogen-Water Properties at 298 K\n", "Dab2 = .63*math.pow(10,-8); #[m^2/s] diffusion coefficient\n", "#Table A.8 Hydrogen-iron Properties at 293 K\n", "Dab3 = .26*math.pow(10,-12); #[m^2/s] diffusion coefficient\n", "#Table A.4 Air properties at 293 K\n", "a1 = 21.6*math.pow(10,-6); #[m^2/s] Thermal Diffusivity\n", "#Table A.6 Water properties at 293 K\n", "k = .603 \t\t\t\t;#[W/m.K] conductivity\n", "rho = 998 \t\t\t\t;#[kg/m^3] Density\n", "cp = 4182 \t\t\t\t;#[J/kg] specific Heat\n", "#Table A.1 Iron Properties at 300 K\n", "a3 = 23.1 * math.pow(10,-6); #[m^2/s]\n", "#calculations\n", "\n", "#Equation 14.14\n", "#Hydrogen-air Mixture\n", "DabT1 = Dab1*math.pow(T/298.,1.5);# [m^2/s] mass diffusivity\n", "J1 = -DabT1*1; \t\t#[kmol/s.m^2] Total molar concentration\n", "j1 = Ma*J1; \t\t#[kg/s.m^2] mass Flux of Hydrogen\n", "Le1 = a1/DabT1; \t# Lewis Number Equation 6.50\n", "\n", "#Hydrogen-water Mixture\n", "DabT2 = Dab2*math.pow(T/298.,1.5);# [m^2/s] mass diffusivity\n", "a2 = k/(rho*cp) \t;#[m^2/s] thermal diffusivity \n", "J2 = -DabT2*1 \t;#[kmol/s.m^2] Total molar concentration\n", "j2 = Ma*J2 \t;#[kg/s.m^2] mass Flux of Hydrogen\n", "Le2 = a2/DabT2 \t;# Lewis Number Equation 6.50\n", "\n", "#Hydrogen-iron Mixture\n", "DabT3 = Dab3*math.pow(T/298.,1.5);# [m^2/s] mass diffusivity\n", "J3 = -DabT3*1; \t#[kmol/s.m^2] Total molar concentration\n", "j3 = Ma*J3; \t#[kg/s.m^2] mass Flux of Hydrogen\n", "Le3 = a3/DabT3 \t;# Lewis Number Equation 6.50\n", "#results\n", "\n", "print '%s %.1e' %('a (m^2/s) in 1 = ',a1)\n", "print '%s %.1e' %('\\n a (m^2/s) in 2 = ',a2)\n", "print '%s %.1e' %('\\na (m^2/s) in 3 = ',a3)\n", "print '%s %.1e' %('\\nDab (m^2/s) in 1 = ',DabT1)\n", "print '%s %.1e' %('\\n Dab (m^2/s) in 2 = ',DabT2)\n", "print '%s %.1e' %('\\n Dab (m^2/s) in 3 = ',DabT3)\n", "print '%s %.1e' %('\\n Le in 1 = ',Le1)\n", "print '%s %.1e' %('\\n Le in 2 = ',Le2)\n", "print '%s %.1e' %('\\n Le in 3 = ',Le3)\n", "print '%s %.1e' %('\\n ja (kg/s.m^2) in 1 = ',j1)\n", "print '%s %.1e' %('\\n ja (kg/s.m^2) in 2 = ',j2)\n", "print '%s %.1e' %('\\n ja (kg/s.m^2) in 3 = ',j3)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a (m^2/s) in 1 = 2.2e-05\n", "\n", " a (m^2/s) in 2 = 1.4e-07\n", "\n", "a (m^2/s) in 3 = 2.3e-05\n", "\n", "Dab (m^2/s) in 1 = 4.0e-05\n", "\n", " Dab (m^2/s) in 2 = 6.1e-09\n", "\n", " Dab (m^2/s) in 3 = 2.5e-13\n", "\n", " Le in 1 = 5.4e-01\n", "\n", " Le in 2 = 2.4e+01\n", "\n", " Le in 3 = 9.1e+07\n", "\n", " ja (kg/s.m^2) in 1 = -8.0e-05\n", "\n", " ja (kg/s.m^2) in 2 = -1.2e-08\n", "\n", " ja (kg/s.m^2) in 3 = -5.1e-13\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.2 Page 898" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Initialization\n", "\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import matplotlib\n", "from matplotlib import pyplot\n", "\n", "T = 298 \t\t\t;#[K] Temperature\n", "D = 10*math.pow(10,-6) \t;#[m]\n", "L = 100*math.pow(10,-6); #[m]\n", "H = .5 \t\t\t;# Moist Air Humidity\n", "p = 1.01325 \t\t\t;#[bar]\n", "#Table A.6 Saturated Water vapor Properties at 298 K\n", "psat = .03165; \t#[bar] saturated Pressure\n", "#Table A.8 Water vapor-air Properties at 298 K\n", "Dab = .26*math.pow(10,-4); #[m^2/s] diffusion coefficient\n", "#calculations\n", "\n", "C = p/(8.314/100. *298) ;#Total Concentration\n", "#From section 6.7.2, the mole fraction at x = 0 is\n", "xa0 = psat/p;\n", "#the mole fraction at x = L is\n", "xaL = H*psat/p;\n", "\n", "#Evaporation rate per pore Using Equation 14.41 with advection\n", "N = (math.pi*D*D)*C*Dab/(4*L)*2.303*math.log10((1-xaL)/(1-xa0)) ;#[kmol/s]\n", "\n", "#Neglecting effects of molar averaged velocity Equation 14.32\n", "#Species transfer rate per pore\n", "Nh = (math.pi*D*D)*C*Dab/(4*L)*(xa0-xaL) ;#[kmol/s]\n", "#results\n", "\n", "print '%s %.2e %s' %('\\n Evaporation rate per pore Without advection effects',Nh,'kmol/s')\n", "print '%s %.2e %s' %('and With Advection effects',N,'kmol/s')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Evaporation rate per pore Without advection effects 1.30e-14 kmol/s\n", "and With Advection effects 1.34e-14 kmol/s\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.3 Page 898" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Initialization\n", "\n", "import math\n", "\n", "D = .005 \t\t\t\t\t\t;#[m] Diameter\n", "L = 50*math.pow(10,-6); \t#[m] Length\n", "h = .003 \t\t\t\t;#[m] Depth\n", "Dab = 6*math.pow(10,-14) \t;#[m^2/s] Diffusion coefficient\n", "Cas1 = 4.5*math.pow(10,-3) \t;#[kmol/m^3] Molar concentrations of water vapor at outer surface\n", "Cas2 = 0.5*math.pow(10,-3) \t;#[kmol/m^3] Molar concentrations of water vapor at inner surface\n", "#calculations\n", "\n", "#Transfer Rate through cylindrical wall Equation 14.54\n", "Na = Dab/L*(math.pi*D*D/4. + math.pi*D*h)*(Cas1-Cas2); #[kmol/s]\n", "#results\n", "\n", "print '%s %.2e %s' %('\\n Rate of water vapor molar diffusive ttansfer through the trough wall ',Na,'kmol/s');\n", "#END" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Rate of water vapor molar diffusive ttansfer through the trough wall 3.20e-16 kmol/s\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.4 Page 902" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Initialization\n", "\n", "import math\n", "\n", "D = .2 \t\t\t;#[m] Diameter\n", "L = 2*math.pow(10,-3) ;#[m] Thickness\n", "p = 4 \t\t\t;#[bars] Helium Pressure\n", "T = 20+273. \t\t\t;#[K] Temperature\n", "#Table A.8 helium-fused silica (293K) Page 952\n", "Dab = .4*math.pow(10,-13)\t;#[m^2/s] Diffusion coefficient\n", "#Table A.10 helium-fused silica (293K)\n", "S = .45*math.pow(10,-3)\t\t;#[kmol/m^3.bar] Solubility\n", "#calculations\n", "\n", "# By applying the species conservation Equation 14.43 and 14.62\n", "dpt = -6*(.08314)*T*(Dab)*S*p/(L*D);\n", "\n", "#results\n", "print '%s %.2e %s' %('\\n The rate of change of the helium pressure dp/dt',dpt,' bar/s');\n", "#END" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " The rate of change of the helium pressure dp/dt -2.63e-11 bar/s\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.5 Page 904" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Initialization\n", "\n", "import math\n", "\n", "Dab = 8.7*math.pow(10,-8) ;#[m^2/s] Diffusion coefficient\n", "Sab = 1.5*math.pow(10,-3) ;#[kmol/m^3.bar] Solubility\n", "L = .0003 \t\t\t;#[m] thickness of bar\n", "p1 = 3 \t\t\t;#[bar] pressure on one side\n", "p2 = 1 \t\t\t;#[bar] pressure on other side\n", "Ma = 2 \t\t\t;#[kg/mol] molecular mass of Hydrogen\n", "#calculations\n", "\n", "#Surface molar concentrations of hydrogen from Equation 14.62\n", "Ca1 = Sab*p1 \t\t\t\t; #[kmol/m^3]\n", "Ca2 = Sab*p2 \t\t\t\t; #[kmol/m^3]\n", "#From equation 14.42 to 14.53 for obtaining mass flux\n", "N = Dab/L*(Ca1-Ca2) ; \t#[kmol/s.m^2]\n", "n = Ma*N ; \t#[kg/s.m^2] on Mass basis\n", "#results\n", "\n", "print '%s %.2e %s' %('\\n The Hydrogen mass diffusive flux n =',n,' (kg/s.m^2)');\n", "#END" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " The Hydrogen mass diffusive flux n = 1.74e-06 (kg/s.m^2)\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.6 Page 909 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Initialization\n", "\n", "import math\n", "\n", "Dab = 2*math.pow(10,-12) \t;#[m^2/s] Diffusion coefficient\n", "Ca0 = 4*math.pow(10,-3) \t\t;#[kmol/m^3] Fixed Concentration of medication\n", "Na = -.2*math.pow(10,-3) \t\t;#[kmol/m^3.s] Minimum consumption rate of antibiotic\n", "k1 = .1 \t\t\t\t\t;#[s^-1] Reaction Coefficient\n", "#calculations\n", "\n", "#For firsst order kinetic reaction Equation 14.74\n", "m = math.pow((k1/Dab),.5);\n", "L = math.acosh(-k1*Ca0/Na) /m;\n", "#results\n", "\n", "print '%s %.1f %s' %('\\n Maximum Thickness of a bacteria laden biofilm, that may be siccessfully treated is ',L*math.pow(10,6), 'mu-m');\n", "#END" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Maximum Thickness of a bacteria laden biofilm, that may be siccessfully treated is 5.9 mu-m\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.7 Page 913" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Initialization\n", "\n", "import math\n", "\n", "Dap = .1*math.pow(10,-12) ;#[m^2/s] Diffusion coefficient of medication with patch\n", "Das = .2*math.pow(10,-12) ;#[m^2/s] Diffusion coefficient of medication with skin\n", "L = .05 \t\t\t;#[m] patch Length\n", "rhop = 100 \t\t\t;#[kg/m^3] Density of medication on patch\n", "rho2 = 0 \t\t\t;#[kg/m^3] Density of medication on skin\n", "K = .5 \t\t\t;#Partition Coefficient\n", "t = 3600*24*7 \t\t\t;#[s] Treatment time\n", "#calculations\n", "\n", "#Applying Conservation of species equation 14.47b\n", "#By analogy to equation 5.62, 5.26 and 5.58\n", "D = 2*rhop*L*L/(math.sqrt(math.pi))*math.sqrt(Das*Dap*t)/(math.sqrt(Das)+math.sqrt(Dap)/K);\n", "#results\n", "\n", "print '%s %.1f %s' %('\\n Total dosage of medicine delivered to the patient over a one-week time period is',D*math.pow(10,6) ,'mg');" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Total dosage of medicine delivered to the patient over a one-week time period is 28.7 mg\n" ] } ], "prompt_number": 10 } ], "metadata": {} } ] }