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diff --git a/Elements_Of_Mass_Transfer_Part_1/ch2.ipynb b/Elements_Of_Mass_Transfer_Part_1/ch2.ipynb new file mode 100644 index 00000000..b72246a4 --- /dev/null +++ b/Elements_Of_Mass_Transfer_Part_1/ch2.ipynb @@ -0,0 +1,1682 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : Diffusion" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.1.a " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusivity of nitrogen carbondioxide mixture\n", + "\n", + "# let A denote nitrogen and B denote carboondioxide\n", + "\n", + "# variable declaration \n", + "rA=.3798;\n", + "rB=.3941;\n", + "\n", + "# Calculation\n", + "rAB=(rA+rB)/2; #molecular seperation at collision \n", + "ebyk_A=71.4;\n", + "ebyk_B=195.2;\n", + "ebyk_AB=(ebyk_A/ebyk_B)**.5; #energy of molecular attraction\n", + "pt=1.013*10**5; #absolute total pressure in pascal\n", + "T=298; #absolute temperature in kelvin\n", + "s=T/ebyk_AB; #collision function\n", + " #from chart f(T/ebyk_AB) = 0.5 let it be = x\n", + "x=.5; #collision function\n", + "MA=28; #molecular weight of nitrogen\n", + "MB=44; #molecular weight of carbondioxide\n", + "Mnew=((1./MA)+(1./MB))**.5;\n", + "Dab= 10**-4*(1.084-.249*(Mnew))*T**1.5*((Mnew))/(pt*x*rAB**2);\n", + "\n", + "# Result\n", + "print \" the diffisivity of nitrogen-carbondioxide is :%f *10**-5 m**2/s\"%(Dab/10**-5)\n", + "\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the diffisivity of nitrogen-carbondioxide is :1.678856 *10**-5 m**2/s\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.1.b " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#diffusivity of hydrogen chloride mixture\n", + "# let A denote Hydrogen chloride and B denote air\n", + "\n", + "# variable declaration \n", + "\n", + "#\"part(ii)\"\n", + "rA=.3339;\n", + "rB=.3711;\n", + "\n", + "# Calculation \n", + "\n", + "rAB=(rA+rB)/2; #molecular seperation at collision \n", + "ebyk_A=344.7;\n", + "ebyk_B=78.6;\n", + "ebyk_AB=(ebyk_A/ebyk_B)**.5; #energy of molecular attraction\n", + "pt=200.*10**3; #absolute total pressure in pascal\n", + "T=298.; #absolute temperature in kelvin\n", + "s=T/ebyk_AB; #collision function\n", + " #from chart f(T/ebyk_AB) = 0.62 let it be = x\n", + "x=0.62; #collision function\n", + "MA=36.5; #molecular weight of hydrogen chloride\n", + "MB=29; #molecular weight of air\n", + "Mnew=((1./MA)+(1./MB))**.5;\n", + "Dab=10.**-4*(1.084-.249*(Mnew))*T**1.5*((Mnew))/(pt*x*rAB**2);\n", + "\n", + " # Result\n", + "print \"\\n the diffisivity of hydrogen chloride-air is :%f *10**-6 m**2/s\"%(Dab/10**-6)\n", + "\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the diffisivity of hydrogen chloride-air is :8.488596 *10**-6 m**2/s\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.2 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# the diffusivity of isoamyl alcohol\n", + "#kopp's law is valid\n", + "\n", + "# variable declaration \n", + "\n", + "u=1.145*10**-3; #viscosity of water1.145cp\n", + "v_a=5*.0148+12*.0037+1*.0074; #by kopp's law\n", + "t=288; #temperature of water in kelvin\n", + "MB=18; #molecular weight of water\n", + "phi=2.26; #association parameter for solvent-water\n", + "\n", + "# Calculation\n", + "D_ab=(117.3*10**-18)*((phi*MB)**.5)*(t)/(u*(v_a)**.6);\n", + "\n", + " # Result\n", + "print \"\\n the diffusivity of isoamyl alcohol is :%f *10**-9 m**2/s\"%(D_ab/10**-9)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the diffusivity of isoamyl alcohol is :0.652780 *10**-9 m**2/s\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.3 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusivity of ccl4 through oxygen\n", + "\n", + "# variable declaration \n", + "\n", + "pa1=(33./760)*1.013*10**5; #vapour pressure of ccl4 at 273 in pascal\n", + "pa2=0;\n", + "d=1.59; \n", + "import math #density of liquid ccl4 in g/cm**3\n", + "#considering o2 to be non diffusing and with \n", + "T=273.; #temperature in kelvin\n", + "pt=(755./780)*1.013*10**5; #total pressure in pascal\n", + "z=.171; #thickness of film\n", + "a=.82*10**-4; #cross-sectional area of cell in m**2\n", + "v=.0208; #volume of ccl4 evaporated \n", + "t=10.; #time of evaporation\n", + "MB=154.; \n", + "\n", + "# Calculation\n", + "#molecular wght of ccl4\n", + "rate=v*d/(MB*t); #.0208cc of ccl4 is evaporating in 10hrs \n", + "Na=rate*10.**-3/(3600.*a); #flux in kmol/m**2*S\n", + "\n", + "D_ab=Na*z*8314*273./(pt*(math.log((pt-pa2)/(pt-pa1)))); #molecular diffusivity in m**2/s\n", + "\n", + " # Result\n", + "print \"the diffusivity of ccl4 through oxygen:%.2f *10**-6 m**2/s\"%(D_ab/10.**-6)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the diffusivity of ccl4 through oxygen:6.27 *10**-6 m**2/s\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.4 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate at which crystal dissolves\n", + "\n", + "\n", + "# variable declaration \n", + "\n", + "z=.0305*10**-3; #wall thickness sorrounding the crystal\n", + "x1=0.0229;\n", + "w1=160.; #molecular weight of copper sulphate\n", + "w2=18.; #molecular weight of water\n", + "Dab=7.29*10**-10; #diffusivity of copper sulphatein m**2/s\n", + " #av=d/m\n", + " \n", + "# Calculation \n", + "Mavg=x1*w1+(1-x1)*w2; #average molecular wght of solution\n", + "d1=1193; #density of copper sulphate solution\n", + "av1=d1/Mavg; #value of (d/m) of copper solution\n", + "\n", + " #for pure water\n", + "d2=1000.; #density of water\n", + "m2=18.; #molecular wght of water\n", + "av2=d2/m2; #value of (d/m) of water\n", + "allavg=(av1+av2)/2.; #average value of d/m\n", + "xa2=0;\n", + "import math\n", + "\n", + "# Result\n", + "Na=Dab*(allavg)*math.log((1-xa2)/(1-x1))/z; #flux of cuso4 from crystal surface to bulk solution\n", + "print \" the rate at which crystal dissolves :%f *10**-5 kmol/m**2*s\"%(Na/10**-5)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the rate at which crystal dissolves :3.092260 *10**-5 kmol/m**2*s\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.5.a" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate of diffusion of alcohol water vapour mixture\n", + "#position 1 moles molefraction\n", + "# air 80 0.8\n", + "# water 20 0.2\n", + "\n", + "#position 2 moles molefraction\n", + "# air 10 0.1\n", + "# water 90 0.9\n", + "\n", + "# variable declaration \n", + "\n", + "ya1=0.8;\n", + "ya2=0.1;\n", + "T=(273+35); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=0.3*10**-3; #gas film thickness in m\n", + "Dab=.18*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314; #universal gas constant\n", + "\n", + "# Calculation\n", + "Na=Dab*pt*(ya1-ya2)/(z*R*T) #diffusion flux in kmol/m**2*s\n", + "rate=Na*100*10**-4*3600*46; #since molecular weight of mixture is 46\n", + "\n", + "# Result\n", + "print \"\\n rate of diffusion of alcohol-water vapour :%f kg/hr \"%rate\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " rate of diffusion of alcohol-water vapour :2.751429 kg/hr \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.5.b" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate of diffusion if water layer is stagnant\n", + "\n", + "# variable declaration \n", + "\n", + "ya1=0.8;\n", + "ya2=0.1;\n", + "T=(273+35); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=0.3*10**-3; #gas film thickness in m\n", + "Dab=.18*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314; #universal gas constant\n", + "import math\n", + "#diffusion through stagnant film \n", + "\n", + "# Calculation\n", + "Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "rate=Na*100*10**-4*3600*46; #since molecular weight of mixture is 46\n", + "\n", + "# Result\n", + "print \"\\n rate of diffusion if water layer is stagnant :%f *10**-3 kg/s \"%(rate/(3600*10**-3))\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " rate of diffusion if water layer is stagnant :1.642207 *10**-3 kg/s \n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.6 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate of loss of hydrogen\n", + "\n", + "# variable declaration \n", + "T=298.; #temperature in kelvin\n", + "pt=1.*1.013*10**5; #total pressure in pascal\n", + "ID=25.*10**-3; #internal diameter in m of unvulcanised rubber in m\n", + "OD=50.*10**-3; #internal diameter in m of unvulcanised rubber in m\n", + "Ca1=2.37*10**-3; #conc. of hydrogen at the inner surface of the pipe in kmol/m**3\n", + "Ca2=0; #conc. of hydrogen at 2\n", + "Dab=1.8*10**-10; #diffusion coefficient in cm**2/s\n", + "l=2; #length of pipe in m\n", + "import math\n", + "\n", + "# Calculation \n", + "# Va=Da*Sa*(pa1-pa2)/z;\n", + "z=(50-25)*10**-3/2.; #wall thickness in m\n", + "Va=Dab*(Ca1-Ca2)/z; #diffusion through a flat slab of thickness z \n", + "Sa=2*3.14*l*(OD-ID)/(2*math.log(OD/ID)); #average mass transfer area of \n", + "rate=Va*Sa; #rate of loss of hydrogen by diffusion\n", + "\n", + "\n", + "# Result\n", + "\n", + "print \"\\n rate of loss hydrogen by diffusion through a pipe of 2m length :%f*10**-12kmol/s\"%(rate/10.**-12)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " rate of loss hydrogen by diffusion through a pipe of 2m length :7.730099*10**-12kmol/s\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.7 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# ammonia diffusion through nitrogen\n", + "\n", + "# variable declaration \n", + "pa1=(1.5)*10**4; #vapour pressure of ammonia at pt.1 \n", + "pa2=(0.5)*10**4; #vapour pressure of ammonia at pt.2\n", + "Dab=2.3*10**-5 #molecular diffusivity in m**2/s\n", + "z=0.15; #diffusion path in m\n", + "R=8314.; #universal gas constant \n", + " #ammonia diffuses through nitrogen under equimolar counter diffusion\n", + "T=298.; #temperature in kelvin\n", + "pt=1.013*10**5; #total pressure in pascal\n", + "\n", + "# Calculation\n", + "Na=Dab*(pa1-pa2)/(z*R*T); #flux in kmol/m**2*S\n", + "\n", + "# Result\n", + "print \"the ammonia diffusion through nitrogen under equimolar \\\n", + " counter diffusion:%f *10**-7 kmol/m**2*s\"%(Na/10**-7);\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the ammonia diffusion through nitrogen under equimolar counter diffusion:6.188855 *10**-7 kmol/m**2*s\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate at which crystal dissolves\n", + "#position 1 moles molefraction weight\n", + "# ethanol 0.1478 0.02775 6.80\n", + "# water 5.18 0.9722 93.20\n", + "\n", + "import math\n", + "#position 2 moles molefraction weight\n", + "# ethanol 0.235 0.0453 10.8\n", + "# water 4.96 0.9547 89.20\n", + "\n", + "# variable declaration \n", + "\n", + "z=0.4*10**-2; #film thickness sorrounding the crystal\n", + "xa1=0.0453; #mole fraction of ethanol at pos.2\n", + "xa2=0.02775; #mole fraction of ethanol at pos.1\n", + "w1=46; #molecular weight of ethanol\n", + "w2=18; #molecular weight of water\n", + "Dab=74*10**-5*10**-4; #diffusivity of ethanol water sol.in m**2/s\n", + " #av=d/m\n", + " \n", + "# Calculation \n", + "Mavg1=xa2*w1+(1-xa2)*w2; #average molecular wght of solution at pos 1\n", + "d1=0.9881*10**3; # density of 6.8 wt% solution\n", + "av1=d1/Mavg1; #value of (d/m) of copper solution\n", + "\n", + " #for pure water\n", + "d2=972.8; # density of 10.8 wt% solution\n", + "Mavg2=xa1*w1+(1-xa1)*w2; #average molecular wght of solution at pos.2\n", + "av2=d2/Mavg2; #value of (d/m) of water\n", + "\n", + "allavg=(av1+av2)/2; #average value of d/m\n", + "Na=Dab*(allavg)*math.log((1-xa2)/(1-xa1))/z; #steady state flux in kmol/m**2*s of ethanol water sol.\n", + "\n", + " # Result\n", + "print \"\\n the rate at which crystal dissolves :%f *10**-5 kmol/m**2*s\"%(Na/10**-5)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the rate at which crystal dissolves :1.737360 *10**-5 kmol/m**2*s\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion rate of acetic acid\n", + "#position 1 kmoles molefraction weight\n", + "# acetic acid 0.167 0.0323 10\n", + "# water 5 0.9677 90\n", + "\n", + "#position 2 kmoles molefraction weight\n", + "# aceitic acid 0.067 0.0124 4\n", + "# water 5.33 0.9876 96\n", + "import math\n", + "# variable declaration \n", + "\n", + " #basis : 100kg of mixture\n", + "z=2*10**-3; #film thickness sorrounding the water\n", + "xa1=0.0323; #mole fraction of ethanol at pos.2\n", + "xa2=0.0124; #mole fraction of ethanol at pos.1\n", + "w1=60.; #molecular weight of acetic acid\n", + "w2=18.; #molecular weight of water\n", + "Dab=0.000095; #diffusivity of acetic water sol.in m**2/s\n", + " #av=d/m\n", + "# Calculation \n", + "Mavg1=xa1*w1+(1-xa1)*w2; #average molecular wght of solution at pos 1\n", + "d1=1013.; # density of 10 % acid\n", + "av1=d1/Mavg1; #value of (d/m) of copper solution\n", + "\n", + " #for pure water\n", + "d2=1004; #density of 4% acid\n", + "Mavg2=xa2*w1+(1-xa2)*w2; #average molecular wght of solution at pos.2\n", + "av2=d2/Mavg2; #value of (d/m) of water\n", + "\n", + "allavg=(av1+av2)/2.; #average value of d/m\n", + " #assuming water to be non diffusing\n", + "Na=Dab*(allavg)*math.log((1-xa2)/(1-xa1))/z; #diffusion rate of acetic acid aacross film of non diffusing water sol.\n", + "\n", + " # Result\n", + "print \"\\n diffusion rate of acetic acid aacross film of non diffusing water sol. :%f kmol/m**2*s\"%Na\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " diffusion rate of acetic acid aacross film of non diffusing water sol. :0.051508 kmol/m**2*s\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.10.a" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate of mass transfer\n", + "# variable declaration \n", + "\n", + "#part (i)\n", + "r=(50./2)*10**-3; #radius pf circular tube\n", + "pa1=190; #vapour pressure of ammonia at pt.1 \n", + "pa2=95; #vapour pressure of ammonia at pt.2\n", + "Dab=2.1*10**-5 #molecular diffusivity in m**2/s\n", + "z=1;\n", + "R=760*22.414/273; #universal gas constant in mmHg*m**3*K*kmol \n", + " #carbondioxide and oxygen experiences equimolar counter diffusion \n", + "T=298.; #temperature in kelvin\n", + "\n", + "# Calculation \n", + "pt=(10.0/780)*1.013*10**5; #total pressure in pascal\n", + "Na=Dab*(pa1-pa2)/(z*R*T); #flux in kmol/m**2*S\n", + "rate=Na*(3.14*r**2); #rate of mass transfer..(3.14*r**2)-is the area\n", + "\n", + "# Result\n", + "print \"\\n the rate of mass transfer.:%f *10**-10 kmol/s\"%(rate/10**-10)\n", + " \n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the rate of mass transfer.:2.105552 *10**-10 kmol/s\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.10.b" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# partial pressure of co2\n", + "#part (i)\n", + "# variable declaration \n", + "\n", + "r=(50./2)*10**-3; #radius pf circular tube\n", + "pa1=(190.); #vapour pressure of ammonia at pt.1 \n", + "pa2=(95.); #vapour pressure of ammonia at pt.2\n", + "Dab=2.1*10**-5 #molecular diffusivity in m**2/s\n", + "R=760.*22.414/273; #universal gas constant in mmHg*m**3*K*kmol \n", + " #carbondioxide and oxygen experiences equimolar counter diffusion \n", + "T=298.; #temperature in kelvin\n", + "\n", + "# Calculation\n", + "pt=(10./780)*1.013*10**5; #total pressure in pascal\n", + "\n", + " #part (ii)\n", + "#(ya-ya1)/(ya2-ya1)=(z-z1)/(z2-z1);\n", + "z2=1; #diffusion path in m at pos.2\n", + "z1=0; #diffusion path in m at pos.1\n", + "z=.75; #diffusion at general z\n", + "#pa=poly([0],'pa'); #calc. of conc. in gas phase\n", + "x= 118.750000 #roots((pa-pa1)/(pa2-pa1)-(z-z1)/(z2-z1));\n", + "\n", + "# Result\n", + "print \"\\n partial pressure of co2 at o.75m from the end where partial pressure is 190mmhg is:%f mmHg\"%x\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " partial pressure of co2 at o.75m from the end where partial pressure is 190mmhg is:118.750000 mmHg\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.11.a" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion flux if N2 is non diffusing\n", + "\n", + "# variable declaration \n", + "\n", + "ya1=0.2; #initial mole fraction\n", + "ya2=0.1; #final mole fraction\n", + "T=298 #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=0.2*10**-2; #gas film thickness in m\n", + "Dab=.215*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314.; #universal gas constant\n", + "#part (i)when N2 is non diffusing \n", + "import math\n", + "\n", + "# Calculation\n", + "Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "\n", + "\n", + "# Result\n", + "print \" diffusion flux if N2 is non diffusing :%f *10**-5 kmol/m**2*s \"%(Na/10**-5);\n", + "\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " diffusion flux if N2 is non diffusing :5.176955 *10**-5 kmol/m**2*s \n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.11.b" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion flux of oxygen\n", + "\n", + "# variable declaration \n", + "\n", + "ya1=0.2;\n", + "ya2=0.1;\n", + "T=(298); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=0.2*10**-2; #gas film thickness in m\n", + "Dab=.215*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314; #universal gas constant\n", + "\n", + "#part (ii) equimolar counter diffusion\n", + "# Calculation\n", + "Na=Dab*pt*(ya1-ya2)/(z*R*T) #diffusion flux in kmol/m**2*s\n", + "\n", + " # Result\n", + "print \" diffusion flux of oxygen during equimolar counter-diffusion :%f *10**-5 kmol/m**2*s \"%(Na/10**-5)\n", + "\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " diffusion flux of oxygen during equimolar counter-diffusion :4.395331 *10**-5 kmol/m**2*s \n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion flux through inert air\n", + "\n", + "# variable declaration \n", + "# ammonia diffusing through inert air and air is non-diffusing\n", + "ya1=0.1;\n", + "ya2=0;\n", + "T=293. #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=0.2*10**-2; #gas film thickness in m\n", + "Dab=.185*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314.; #universal gas constant\n", + " #part (i)when air is assumed to be stagnant and non-diffusing \n", + "import math\n", + "\n", + "# Calculation\n", + "Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "mw=17; #molecular weight of ammonia\n", + "massflux=Na*mw; #mass flux of given NH3\n", + "print \" diffusion flux when total presssure is 1atm and air \\\n", + " is non-diffusing :%f *10**-4 kg/m**2*s \"%(massflux/10**-4);\n", + " #part (ii) when pressure is increased to 10atm\n", + "\n", + "#Dab_1/Dab_2=pt_2/pt_1\n", + "pt_2=10; #final pressure in atm\n", + "pt_1=1; #initially pressure was 1atm\n", + "Dab_1=.185; #initially diffusion coefficient was.185\n", + "Dab_2=Dab_1*pt_1/pt_2; #for gases Dab is proportional to 1/pt\n", + "Dab=Dab_2*10**-4; #new diffusion coefficient \n", + "pt=pt_2*1.013*10**5; #new total pressure\n", + "Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "\n", + "# Result\n", + "print \"diffusion flux when pressure is increased to 10atm :%f *10**-5 kmol/m**2*s \"%(Na/10**-5);\n", + "print \"so the rate of diffusion remains same on increasing the pressure\"\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " diffusion flux when total presssure is 1atm and air is non-diffusing :6.889701 *10**-4 kg/m**2*s \n", + "diffusion flux when pressure is increased to 10atm :4.052765 *10**-5 kmol/m**2*s \n", + "so the rate of diffusion remains same on increasing the pressure\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion rate of acetic acid\n", + "#position 1 moles molefraction weight\n", + "# acetic acid 0.15 0.0288 9\n", + "# water 5 0.9712 91\n", + "\n", + "#position 2 moles molefraction weight\n", + "# aceitic acid 0.05 0.0092 4\n", + "# water 5.389 0.9908 96\n", + "# variable declaration \n", + "T=290.0; #temperature in kelvin\n", + "z=2*10**-3; #film thickness sorrounding the water\n", + "xa2=0.0092; #mole fraction of ethanol at pos.2\n", + "xa1=0.0288; #mole fraction of ethanol at pos.1\n", + "w1=60.0; #molecular weight of acetic acid\n", + "w2=18.0; #molecular weight of water\n", + "Dab=0.95*10**-9; #diffusivity of acetic water sol.in m**2/s\n", + " #av=d/m\n", + "# Calculation \n", + "Mavg1=xa1*w1+(1-xa1)*w2; #average molecular wght of solution at pos 1\n", + "d1=1012.0; # density of 10 % acid\n", + "av1=d1/Mavg1; #value of (d/m) of copper solution\n", + "\n", + " #for position 2\n", + "d2=1003.0; #density of 4% acid\n", + "Mavg2=xa2*w1+(1-xa2)*w2; #average molecular wght of solution at pos.2\n", + "av2=d2/Mavg2; #value of (d/m) of water\n", + "\n", + "allavg=(av1+av2)/2; #average value of d/m\n", + " \n", + " #assuming water to be non diffusing\n", + "import math \n", + "Na=Dab*(allavg)*math.log((1-xa2)/(1-xa1))/z; #diffusion rate of acetic acid aacross film of non diffusing water sol.\n", + "\n", + "# Result\n", + "print \"diffusion rate of acetic acid aacross film of non diffusing water sol. :%f *10**-7 kmol/m**2*s\"%(Na/10.0**-7)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "diffusion rate of acetic acid aacross film of non diffusing water sol. :5.088553 *10**-7 kmol/m**2*s\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion flux of nitrogen\n", + "\n", + "# variable declaration \n", + "\n", + "ya1=0.2; #molefraction at pos.1\n", + "ya2=0.1; #molefraction at pos.2\n", + "T=(293); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=0.2*10**-2; #gas film thickness in m\n", + "Dab=.206*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314; #universal gas constant\n", + " #for ideal gases volume fraction =mole fraction\n", + "#part (i)when N2 is non diffusing \n", + "import math\n", + "\n", + "# Calculation\n", + "Na=Dab*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "print \"\\n diffusion flux if N2 is non diffusing :%f *10**-5 kmol/m**2*s \"%(Na/10**-5);\n", + "#part (ii) equimolar counter diffusion\n", + "\n", + "Na=Dab*pt*(ya1-ya2)/(z*R*T) #diffusion flux in kmol/m**2*s\n", + "\n", + "# Result\n", + "print \"\\n diffusion flux of nitrogen during equimolar counter-diffusion :%f *10**-5 kmol/m**2*s \"%(Na/10**-5)\n", + "\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " diffusion flux if N2 is non diffusing :5.044891 *10**-5 kmol/m**2*s \n", + "\n", + " diffusion flux of nitrogen during equimolar counter-diffusion :4.283207 *10**-5 kmol/m**2*s \n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion rate of loss of benzene\n", + "\n", + "# variable declaration \n", + "\n", + "pa1=0.2*10**5; #partial pressure at pos.1\n", + "pa2=0; #partial pressure at pos.2\n", + "r=10./2; #radius of tank in which benzene is stored\n", + "T=298.; #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=10*10**-3; #gas film thickness in m\n", + "Dab=.02/3600; #diffusion coefficient in m**2/s\n", + "R=8314.; #universal gas constant\n", + " #benzene is stored in atank of dia 10m\n", + "#part (i)when air is assumed to be stagnant\n", + "import math\n", + "\n", + "# Calculation\n", + "Na=Dab*pt*math.log((pt-pa2)/(pt-pa1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "rate=Na*(3.14*r**2); #rate of loss of benzene if air is stagnant\n", + "\n", + " # Result\n", + "print \"diffusion rate of loss of benzene :%f *10**-4 kmol/s \"%(rate/10**-4);\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "diffusion rate of loss of benzene :3.921799 *10**-4 kmol/s \n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate of diffusion of alcohol water vapour mixture\n", + "\n", + "# variable declaration \n", + "\n", + "ya2=0.1; #molefraction at pos.2\n", + "ya1=0.8; #molefraction at pos.1\n", + "T=(370.); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=0.1*10**-3; #gas film thickness in m\n", + "Dab=.15*10**-2; #diffusion coefficient in m**2/s\n", + "R=8314; #universal gas constant\n", + "Area=10; #area of the film is 10m**2\n", + "\n", + " #alcohol is being absorbed infrom amixture of alcohol vapour and water vapour by means of non-volatile solvent in which alcohol is soluble bt water is not \n", + " #for gase Dab=T**3/2\n", + " #Dab1/Dab2=(T1/T2)**3/2\n", + "import math\n", + "T2=370.; #final temperature in kelvin \n", + "T1=298.; #initial temperature in kelvin\n", + "\n", + "# Calculation\n", + "Dab1=.15*10**-2; #initial diffusion coefficient \n", + "Dab2=((T2/T1)**(3./2))*Dab1; #final diffusion coefficient\n", + "Na=Dab2*pt*math.log((1-ya2)/(1-ya1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "rate=Na*3600*46*Area; #rate of diffusion of alcohol-water vapour in kg/hour\n", + "\n", + " # Result\n", + "print \" rate of diffusion of alcohol-water vapour :%f *10**6 kg/hour \"%(rate/10**6)\n", + "\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " rate of diffusion of alcohol-water vapour :1.702149 *10**6 kg/hour \n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion rate of ammonia\n", + "\n", + "# variable declaration \n", + "\n", + "ya2=0; #molefraction at pos.2\n", + "ya1=0.1; #molefraction at pos.1\n", + "T=(273); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=2*10**-3; #gas film thickness in m\n", + "Dab=.198*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314; #universal gas constant\n", + " #ammonia is diffusing through an inert film 2mm thick \n", + "\n", + " #for gase Dab=T**3/2\n", + " #Dab1/Dab2=(T1/T2)**3/2\n", + "import math\n", + "T2=293; #final temperature in kelvin \n", + "T1=273; #initial temperature in kelvin\n", + "\n", + "\n", + "# Calculation\n", + "Dab1=0.198*10**-4; #initial diffusion coefficient \n", + "Dab2=((T2/T1)**(3.0/2))*Dab1; #final diffusion coefficient\n", + "Na=Dab2*pt*math.log((1-ya2)/(1-ya1))/(z*R*T2); #diffusion flux in kmol/m**2*s\n", + "print \" flux of diffusion of ammonia through inert film :%f *10**-5 kmol/m**2*s \"%(Na/10**-5)\n", + "\n", + "#if pressure is also incresed from 1 to 5 atm\n", + " #for gases Dab=(T**3/2)/pt;\n", + " #Dab1/Dab2=(T1/T2)**3/2*(p2/p1)\n", + "T2=293.; #final temperature in kelvin \n", + "T1=273.; #initial temperature in kelvin\n", + "pa2=5.; #final pressure in atm\n", + "pa1=1; #initial pressure in atm \n", + "p=pa2*1.013*10**5;\n", + "Dab1=.198*10**-4; #initial diffusion coefficient\n", + "Dab2=((T2/T1)**(3.0/2))*Dab1*(pa1/pa2); #final diffusion coefficient\n", + "Na=Dab2*p*math.log((1-ya2)/(1-ya1))/(z*R*T2); #diffusion flux in kmol/m**2*s\n", + "\n", + " # Result\n", + "print \"flux of diffusion of ammonia if temp. is 20 and pressure is 5 atm :%f*10**-5 kmol/m**2*s \"%(Na/10**-5)\n", + "print \"so there is no change in flux when pressure is changed\"\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " flux of diffusion of ammonia through inert film :4.337554 *10**-5 kmol/m**2*s \n", + "flux of diffusion of ammonia if temp. is 20 and pressure is 5 atm :4.822834*10**-5 kmol/m**2*s \n", + "so there is no change in flux when pressure is changed\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#ate of evaporation\n", + "\n", + "# variable declaration \n", + "pa1=0.418*10**5; #partial pressure initially\n", + "pa2=0; #partial pressure of pure air\n", + "r=10/2; #radius of tank in which benzene is stored\n", + "T=(350); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "z=2*10**-3; #gas film thickness in m\n", + "Dab=.2*10**-4; #diffusion coefficient in m**2/s\n", + "R=8314; #universal gas constant\n", + "r=0.2/2; #radius of open bowl is 0.2\n", + "#when air layer is assumed to be stagnant of thickness 2mm\n", + "import math\n", + "\n", + "# Calculation\n", + "\n", + "Na=Dab*pt*math.log((pt-pa2)/(pt-pa1))/(z*R*T); #diffusion flux in kmol/m**2*s\n", + "rate=Na*(3.14*r**2)*18; #rate of loss of evaporation\n", + "\n", + "# Result\n", + "print \"\\n diffusion rate loss of evaporation :%f *10**-4 kg/s \"%(rate/10**-4)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " diffusion rate loss of evaporation :1.046972 *10**-4 kg/s \n" + ] + } + ], + "prompt_number": 54 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusivity of the mixture in stefan tube of toluene in air\n", + "\n", + "# variable declaration \n", + "\n", + "#stefan tube experiment\n", + "import math\n", + "Ml=92.; #molecular weight of toluene\n", + "T=(312.4); #temperature in kelvin\n", + "pt=1*1.013*10**5; #total pressure in pascal\n", + "R=8314.; #universal gas constant\n", + "t=275.*3600; #after 275 hours the level dropped to 80mm from the top\n", + "zo=20.*10**-3; #intially liquid toluene is at 20mm from top\n", + "zt=80.*10**-3; #finally liquid toluene is at 80mm from top\n", + "#air is assumed to be satgnant \n", + "d=850.; #density in kg/m**3\n", + "pa=7.64*10**3; #vapour pressure of toluene in at 39.4degree celcius \n", + "\n", + "# Calculation \n", + "cal=d/Ml; #conc. at length at disxtance l\n", + "ca=pt/(R*T); #total conc. \n", + "xa1=pa/pt; #mole fraction of toluene at pt1 i.e before evaporation \n", + "xb1=1-xa1; #mole fraction of air before evaporation i.e at pt1 \n", + "xb2=1.; #mole fraction of air after evaporation i.e at pt.2\n", + "xa2=0.; #mole fraction of toluene at point 2\n", + "xbm=(xb2-xb1)/(math.log(xb2/xb1));\n", + "#t/(zt-zt0) = (xbm*cal*(zt+zo))/(2*c*(xa1-xa2)*t);\n", + "Dab=(xbm*cal*(zt**2-zo**2))/(2*ca*t*(xa1-xa2));\n", + "\n", + "# Result\n", + "print \"\\n the diffusivity of the mixture in stefan tube of toluene in air is :%f*10**-5 m**2/s\"%(Dab/10**-5)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the diffusivity of the mixture in stefan tube of toluene in air is :0.915437*10**-5 m**2/s\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.20 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusion flux of a mixture of benzene and toluene\n", + "\n", + "#this is the case of equimolar counter diffusion as the latent heat of vaporisation are very close to each other\n", + "# variable declaration \n", + "\n", + "T=(360); #temperature in kelvin\n", + "pt=372.4/760; #total pressure in atm\n", + "R=82.06; #universal gas constant\n", + "Dab=0.0506; #diffusion coefficient in cm**2/s \n", + "z=0.254; #gas layer thickness in cm\n", + "vp=368.0/760; #vapour pressure of toluene in atm\n", + "xtol=.3; #mole fractoin of toluene in atm\n", + "pb1=xtol*vp; #partial pressure of toluene\n", + "#since pb1 is .045263 bt in book it is rounded to 0.145\n", + "\n", + "# Calculation \n", + "pb2=xtol*pt; #parial pressure of toluene in vapour phase\n", + "Na=Dab*(pb1-pb2)/(z*R*T); #diffusion flux \n", + "\n", + "# Result\n", + "print \"\\n the diffusion flux of a mixture of benzene and toluene %f*10**-8 gmol/cm**2*s\\n\"%(Na/10**-8)\n", + "print \"\\nthe negative sign indicates that the toluene is getting transferred from gas phase \\\n", + "to liquid phase(hence the transfer of benzene is from liquid to gas phase)\"\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the diffusion flux of a mixture of benzene and toluene -1.171233*10**-8 gmol/cm**2*s\n", + "\n", + "\n", + "the negative sign indicates that the toluene is getting transferred from gas phase to liquid phase(hence the transfer of benzene is from liquid to gas phase)\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusivity of the mixture in stefan tube\n", + "\n", + "#stefan tube experiment(pseudo steady state diffusion)\n", + "# variable declaration \n", + "\n", + "Ml=92.; #molecular weight of toluene\n", + "T=(303.); #temperature in kelvin\n", + "pt=1.*1.013*10**5; #total pressure in pascal\n", + "R=8314.; #universal gas constant\n", + "t=275.*3600; #after 275 hours the level dropped to 80mm from the top\n", + "zo=20.*10**-3; #intially liquid toluene is at 20mm from top\n", + "zt=77.5*10**-3; #finally liquid toluene is at 80mm from top\n", + " #air is assumed to be satgnant \n", + "import math\n", + "# Calculation \n", + "d=820.; #density in kg/m**3\n", + "pa=(57.0/760)*1.0135*10**5; #vapour pressure of toluene in at 39.4degree celcius \n", + "cal=d/Ml; #conc. at length at disxtance l\n", + "ca=pt/(R*T); #total conc. \n", + "xa1=pa/pt; #mole fraction of toluene at pt1 i.e before evaporation \n", + "xb1=1.-xa1; #mole fraction of air before evaporation i.e at pt1 \n", + "xb2=1.; #mole fraction of air after evaporation i.e at pt.2\n", + "xa2=0; #mole fraction of toluene at point 2\n", + "xbm=(xb2-xb1)/(math.log(xb2/xb1));\n", + "#t/(zt-zt0) = (xbm*cal*(zt+zo))/(2*c*(xa1-xa2)*t);\n", + "Dab=(xbm*cal*(zt**2-zo**2))/(2*ca*t*(xa1-xa2));\n", + "\n", + "# Result\n", + "print \"\\n the diffusivity of the mixture in stefan tube of toluene in air is :%f*10**-5 m**2/s\"%(Dab/10**-5)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the diffusivity of the mixture in stefan tube of toluene in air is :0.804587*10**-5 m**2/s\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusivity of ccl4\n", + "# variable declaration \n", + "#variation in liquid level with respect to time is given below\n", + "%pylab inline\n", + "t=[26,185,456,1336,1958,2810,3829,4822,6385]\n", + "# let Zt-Zo= x;\n", + "x=[.25,1.29,2.32,4.39,5.47,6.70,7.38,9.03,10.48]\n", + "i=0; \n", + "y = [0,0,0,0,0,0,0,0,0] #looping starts\n", + "\n", + "# Calculation \n", + "while(i<9):\n", + " y[i]=t[i]/x[i]; #for calculating the t/Zt-Zo value\n", + " i=i+1;\n", + "from matplotlib.pyplot import *\n", + "plot(x,y,\"o-\");\n", + "show()\n", + "#xtitle(\" Fig.2.2 Example 22 \",\"X--(zi-zo),cm --->\",\"Y-- vs (t/(zi-zo))min/cm ---->\");\n", + "slope=51.4385*60 *10**4; #slope of the curve in 1/sec*m**2\n", + "#slope = Cal *(xblm)/(2*Dab*C*(xa1-xa2))\n", + "d=1540.; #density in kg/m**3\n", + "Ml=154.; #molecular weight of toluene\n", + "Cal=d/Ml ; #conc. at length at disxtance l in mol/m**3\n", + "\n", + "T=(321.); #temperature in kelvin\n", + "pt=1.; #total pressure in atm\n", + "R=82.06; #universal gas constant\n", + "C=pt/(R*T) *10**3; #total conc. in kg mol/m**3\n", + "\n", + "pa=(282./760); #vapour pressure of toluene \n", + "xa1=pa/pt; #mole fraction of toluene at pt1 i.e before evaporation\n", + "xb1=1.-xa1; #mole fraction of air before evaporation i.e at pt1 \n", + "xb2=1.; #mole fraction of air after evaporation i.e at pt.2\n", + "xa2=0.; #mole fraction of toluene at point 2\n", + "xblm=(xb2-xb1)/(math.log(xb2/xb1)); #log mean temp. difference\n", + "Dab = Cal *(xblm)/(2*slope*C*(xa1-xa2)); #diffusivity coefficient\n", + "\n", + "# Result\n", + "print \"\\n the diffusivity of the mixture by winklemann method of toluene in air is :%f*10**-6 m**2/s\"%(Dab/10**-6)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "WARNING: pylab import has clobbered these variables: ['rate', 'draw_if_interactive']\n", + "`%pylab --no-import-all` prevents importing * from pylab and numpy\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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VUAf84cPw5JOQkGB271u3wu9/D337Wl2ZiIj1gjLg6+rgqaegf3+oqjI33/jj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\nImJTCngREZtSwIuI2JQCXkTEpv4/Y/eFADqmRiMAAAAASUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x20a87d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the diffusivity of the mixture by winklemann method of toluene in air is :9.202857*10**-6 m**2/s\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate of transfer of nitrogen and hydrogen\n", + "# variable declaration \n", + "#it is the case of equimolar conter diffusion as the tube is perfectly sealed to two bulbs at the end and the pressure throughout is constant\n", + "d=0.001;\n", + "area=3.14*(d/2)**2; #area of the bulb\n", + "T=298.; #temperature in kelvin \n", + "p=1.013*10**5; #total pressure of both the bulbs\n", + "R=8314.; #universal gas constant\n", + "\n", + "# Calculation \n", + "c=p/(R*T); #total concentration\n", + "Dab=.784*10**-4; #diffusion coefficient in m**2/s\n", + "xa1=0.8; #molefraction of nitrogen gas at the 1 end\n", + "xa2=0.25; #molefraction of nitrogen gas at the 2nd end\n", + "z=.15; #distance between the bulbs\n", + "\n", + " #rate=area*Na;\n", + "rate=area*Dab*c*(xa1-xa2)/z; #rate of transfer of hydrogen and hydrogen\n", + "\n", + "# Result\n", + "print \"\\n the rate of transfer from 1 to 2 of nitrogen and 2 to 1of hydrogen is :%f *10**-11kmol/s\"%(rate/10.0**-11)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the rate of transfer from 1 to 2 of nitrogen and 2 to 1of hydrogen is :0.922657 *10**-11kmol/s\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusivity of methanol in carbon tetrachloride\n", + "# variable declaration \n", + "#using wilke and chang empirical correlation\n", + "#Dab=(117.3*10**-18)*(phi*Mb)**0.5*T/(u*va**0.6);\n", + "\n", + "T=288; #temperature in kelvin\n", + "Mb=32; #molecular weight of methanol\n", + "phi=1.9; #association factor for solvent\n", + "\n", + "# Calculation \n", + "va=(14.8+(4*24.6))*10**-3 #solute(CCl4) volume at normal BP in m**3/kmol\n", + "u=.6*10**-3; #viscosity of solution in kg/m*s\n", + "Dab=(117.3*10**-18)*(phi*Mb)**0.5*T/(u*va**0.6); #diffusion coefficient in m**2/s\n", + "\n", + "# Result\n", + "print \"\\ndiffusivity of methanol in carbon tetrachloride is :%f*10**-9 m**2/s\"%(Dab/10.0**-9)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "diffusivity of methanol in carbon tetrachloride is :1.622494*10**-9 m**2/s\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# diffusivity of methanol in water\n", + "\n", + "#using wilke and chang empirical correlation\n", + "#Dab=(117.3*10**-18)*(phi*Mb)**0.5*T/(u*va**0.6);\n", + "# variable declaration \n", + "T=288.0; #temperature in kelvin\n", + "Mb=18.0; #molecular weight of methanol\n", + "phi=2.26; #association factor for solvent\n", + "\n", + "# Calculation \n", + "va=(2*14.8+(6*3.7)+7.4)*10**-3 #solute(water) volume at normal BP in m**3/kmol\n", + "u=1*10**-3; #viscosity of solution in kg/m*s\n", + "Dab=(117.3*10**-18)*(phi*Mb)**0.5*T/(u*va**0.6); #diffusion coefficient in m**2/s\n", + "\n", + "# Result\n", + "print \"\\ndiffusivity of methanol in water is :%f*10**-9 m**2/s\"%(Dab/10**-9)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "diffusivity of methanol in water is :1.174865*10**-9 m**2/s\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 2.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rate of passage of hydrogen\n", + "# variable declaration \n", + "u=20.*10**-6; #viscosity in Ns/m**2\n", + "pt=2666.; #total pressure in N/m**2\n", + "pa1=pt; #pressure at 1\n", + "pa2=0; #pressure at 2\n", + "mw=32.; #molecular weight of oxygen\n", + "R=8314.; #universal law constant\n", + "T=373.; #temp. in kelvin\n", + "gc=1.; \n", + "\n", + "# Calculation \n", + "l=(3.2*u/pt)*((R*T)/(2*3.14*gc*mw))**0.5;#mean free path\n", + "d=.2*10**-6; #pore diameter\n", + "s=d/l; #value of dia/l\n", + " #hence knudsen diffusion occurs\n", + "Na=0.093*20*273/(760*373*22414*10**-1); #diffusion coefficient in kmol/m**2*s\n", + "Dka=(d/3)*((8*gc*R*T)/(3.14*mw))**0.5;\n", + "l1=Dka*(pa1-pa2)/(R*T*Na); #length of the plate\n", + "\n", + "# Result\n", + "print \"\\n the length of the plate is :%f m \"%l1\n", + "\n", + "\n", + " #for diffusion with hydrogen\n", + "u=8.5*10**-6; #viscosity in Ns/m**2\n", + "pt=1333; #total pressure in N/m**2\n", + "pa1=pt; #pressure at 1\n", + "pa2=0; #pressure at 2\n", + "mw=2; #molecular weight of oxygen\n", + "R=8314; #universal law constant\n", + "T=298; #temp. in kelvin\n", + "gc=1; \n", + "l=(3.2*u/pt)*((R*T)/(2*3.14*gc*mw))**0.5;#mean free path\n", + "d=.2*10**-6; #pore diameter\n", + "s=d/l; #value of dia/l\n", + " #hence knudsen diffusion occurs\n", + "Dka=(d/3)*((8*gc*R*T)/(3.14*mw))**0.5;\n", + "Na=Dka*(pa1-pa2)/(R*T*l1); #diffusion coefficient in kmol/m**2*s\n", + "print \"\\n the diffusion coefficient is :%f *10**-4 kmol/m**2*s\"%(Na/10.0**-6)\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " the length of the plate is :0.035635 m \n", + "\n", + " the diffusion coefficient is :1.788175 *10**-4 kmol/m**2*s\n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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