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diff --git a/Stoichiometry_And_Process_Calculations/ch4.ipynb b/Stoichiometry_And_Process_Calculations/ch4.ipynb new file mode 100755 index 00000000..a8fae4f9 --- /dev/null +++ b/Stoichiometry_And_Process_Calculations/ch4.ipynb @@ -0,0 +1,796 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Ideal Gases and Gas Mixtures" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 page no : 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables \n", + "P1 = 760. #mmHg standard conditions\n", + "T1 = 273.15; #K\n", + "\n", + "# Calculation and Result \n", + "V1 = 22.4143 * 10**-3; #m**3/mol\n", + "R1 = P1 * V1 / T1;\n", + "print \"Gas constant R = %.4e m**3 mmHg / (molK)\"%R1\n", + "P2 = 101325; #N/m**2\n", + "T2 = 273.15; #K\n", + "V2 = 22.4143 * 10**-3; #m**3/mol\n", + "R2 = P2 * V2 / T2; #J/molK\n", + "R3 = R2 / 4.184; #cal/molK\n", + "print \"Gas constant R in MKS system = %.3f cal/molK\"%R3\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Gas constant R = 6.2365e-02 m**3 mmHg / (molK)\n", + "Gas constant R in MKS system = 1.987 cal/molK\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 page no : 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables \n", + "T = 350. #K\n", + "P = 1; #bar\n", + "\n", + "# Calculation \n", + "V1 = 22.4143 * 10**-3; #m**3 (suffix 1 represents at STD)\n", + "P1 = 1.01325; #bar\n", + "T1 = 273.15; #K\n", + "V = P1 * V1 * T/(T1 * P);\n", + "\n", + "# Result \n", + "print \"Molar volume = %.2e m**3/mol\"%V\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molar volume = 2.91e-02 m**3/mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 page no : 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables \n", + "P = 10. #bar oxygen cylinder gas\n", + "T = 300.; #K\n", + "V = 150.; #L\n", + "P1 = 1.01325; #bar ( \\suffix 1 represents at STD)\n", + "T1 = 273.15; #K\n", + "\n", + "# Calculation \n", + "V2 = T1 * P * V /(T * P1); #m**3\n", + "V1 = 22.4143; #m**3/mol\n", + "N = V2 / V1; #mol\n", + "MO2 = 32.;\n", + "m = N * MO2/1000;\n", + "\n", + "# Result \n", + "print \"Mass of oxygen in the cylinder = %.4f kg\"%m\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of oxygen in the cylinder = 1.9243 kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 page no: 81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P = 195. #kPa pressure\n", + "T = 273. #K automobile tyre\n", + "P1 = 250 #kPa pressure\n", + "\n", + "# Calculation \n", + "T1 = P1 * T / P;\n", + "\n", + "# Result\n", + "print \"Maximum temperature to which tyre may be heated = \",T1,\"K\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum temperature to which tyre may be heated = 350.0 K\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 page no : 81\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "# variables \n", + "V = 250. #L carbon dioxide\n", + "T = 300. #K temperature\n", + "V1 = 1000. #L \n", + "P1 = 100. #kPa cylinder\n", + "T1 = 310. #K temperature\n", + "\n", + "# Calculation \n", + "P = T * P1 * V1 /(T1 * V);\n", + "\n", + "# Result\n", + "print \"Original pressure in the cylinder = %.1f kPa\"%P\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Original pressure in the cylinder = 387.1 kPa\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 pageno : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "Vper1 = 70. #% ( 1 = HCl) rubber showed\n", + "Vper2 = 20. #% ( 2 = Cl2) volume\n", + "Vper3 = 10. #% ( 3 = CCl4)\n", + "M1 = 36.45 # molecular weights\n", + "M2 = 70.90\n", + "M3 = 153.8\n", + "\n", + "# Calculation \n", + "m1 = Vper1 * M1\n", + "m2 = Vper2 * M2\n", + "m3 = Vper3 * M3\n", + "mper1 = m1 * 100/(m1+ m2 + m3);\n", + "mper2 = m2 * 100/(m1+ m2 + m3);\n", + "mper3 = m3 * 100/(m1+ m2 + m3);\n", + "\n", + "# Result\n", + "print \" (a) weight percent of HCl= %.2f%%\"%mper1\n", + "print \"weight percent of Cl2 = %.2f %%\"%mper2\n", + "print \"weight percent of CCl4 = %.2f %%\"%mper3\n", + "\n", + "m = (m1 + m2 + m3)/(Vper1 + Vper2 + Vper3);\n", + "print \"(b)average molecular weight = %.3f kg\"%m\n", + "v = 22.4143; #m**3/kmol\n", + "Vtotal = v * (Vper1 + Vper2 + Vper3);\n", + "D = (m1 + m2 + m3)/Vtotal;\n", + "print \"(c)Density at standard condiions = %.4f kg/m**3\"%D\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " (a) weight percent of HCl= 46.33%\n", + "weight percent of Cl2 = 25.75 %\n", + "weight percent of CCl4 = 27.93 %\n", + "(b)average molecular weight = 55.075 kg\n", + "(c)Density at standard condiions = 2.4571 kg/m**3\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 page no : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "per1 = 93. #% ( 1 = methane)\n", + "per2 = 4.5; #% (2 = ethane)\n", + "per3 = 100 - (per1 + per2); #% ( 3 = N2);\n", + "T = 300. #K natural gas \n", + "p = 400. #kPa\n", + "P3 = p * per3 / 100;\n", + "v = 10. #m**3\n", + "V2 = per2 * v / 100;\n", + "M1 = 16.032;\n", + "M2 = 30.048;\n", + "M3 = 28;\n", + "N1 = per1;\n", + "N2 = per2;\n", + "N3 = per3;\n", + "\n", + "# Calculation \n", + "m1 = M1 * N1;\n", + "m2 = M2 * N2;\n", + "m3 = M3 * N3;\n", + "m = m1 + m2 + m3;\n", + "Vstp = 100 * 22.4143 * 10**-3; #m3 at STP\n", + "D = m /(1000 * Vstp);\n", + "Pstp = 101.325; #kPa\n", + "T1 = 273.15; #K\n", + "V = T * Pstp * Vstp / ( T1 * p);\n", + "D1 = m /(1000 * V);\n", + "Mavg = m /100;\n", + "mper1 = m1 * 100 / (m1 + m2 + m3);\n", + "mper2 = m2 * 100 / (m1 + m2 + m3);\n", + "mper3 = m3 * 100 / (m1 + m2 + m3);\n", + "\n", + "# Result\n", + "print \"(a) Partial pressure of nitrogen = \",P3,\"kPa\"\n", + "print \"(b) pure-component volume of ethane = \",V2,\"m**3\"\n", + "print \"(c) Density at standard conditions = %.4f kg/m**3\"%D\n", + "print \"(d) Density at given condition = %.4f kg/m**3\"%D1\n", + "print \"(e) Average molecular weight = %.2f\"%Mavg\n", + "print \"(f) weight percent of Methane = %.2f %%\"%mper1\n", + "print \"weight percent of Ethane = %.2f %%\"%mper2\n", + "print \"weight percent of Nitrogen = %.2f %%\"%mper3\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) Partial pressure of nitrogen = 10.0 kPa\n", + "(b) pure-component volume of ethane = 0.45 m**3\n", + "(c) Density at standard conditions = 0.7567 kg/m**3\n", + "(d) Density at given condition = 2.7200 kg/m**3\n", + "(e) Average molecular weight = 16.96\n", + "(f) weight percent of Methane = 87.90 %\n", + "weight percent of Ethane = 7.97 %\n", + "weight percent of Nitrogen = 4.13 %\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 pageno : 87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "per1 = 20. #% ( 1 = ammonia) \n", + "Vstp = 22.4143; #m**3/kmol\n", + "Pstp = 101.325; #kPa\n", + "Tstp = 273.15; #K\n", + "V1 = 100. #m**3 ammonia air\n", + "P1 = 120. #kPa absorption column\n", + "T1 = 300. #K\n", + "P2 = 100. #kPa gas leaves\n", + "T2 = 280. #K gas leaves\n", + "per2 = 90. #% (absorbed)\n", + "\n", + "# Calculation \n", + "N = V1 * P1 * Tstp / (Vstp * Pstp * T1); #kmol\n", + "Nair = (1 - per1 / 100) * N;\n", + "N1 = per1 * N/100;\n", + "Nabs = per2 * N1 / 100;\n", + "N2 = N1 - Nabs; #leaving\n", + "Ntotal = Nair + N2;\n", + "Vstp1 = Ntotal * Vstp; #m**3\n", + "V2 = Vstp1 * Pstp * T2 / (Tstp * P2);\n", + "\n", + "# Result\n", + "print \"Volume of gas leaving = \",V2, \"m**3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of gas leaving = 91.84 m**3\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 pageno : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "V = 100. #m**3 air\n", + "Ptotal = 100. #kPa air\n", + "Pwater = 4. #kPa pressure\n", + "Pair = Ptotal - Pwater; \n", + "T = 300. #K air\n", + "T1 = 275. #K air cooled\n", + "Vstp = 22.4143; #m**3/kmol\n", + "Tstp = 273.15; #K\n", + "Pstp = 101.325; #kPa\n", + "Pwater1 = 1.8; #kPa\n", + "\n", + "# Calculation \n", + "Pair1 = Ptotal - Pwater1;\n", + "V1 = V * Pair * T1 / ( T * Pair1);\n", + "Nwater = V * Pwater * Tstp/ (Vstp * Pstp * T);\n", + "Nwater1 = V1 * Pwater1 * Tstp/ (Vstp * Pstp * T1);\n", + "m = (Nwater - Nwater1) * 18.02;\n", + "\n", + "# Result\n", + "print \"(a) volume of air after dehumidification = %.2f m**3\"%V1\n", + "print \"(b) Mass of water vapour removed = %.3f kg\"%m\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) volume of air after dehumidification = 89.61 m**3\n", + "(b) Mass of water vapour removed = 1.618 kg\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10 pageno : 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables \n", + "V = 100. #m**3 rate \n", + "P = 600. #kPa gas\n", + "T = 310. #K gas\n", + "per1 = 20. #% ( H2S entering )\n", + "per2 = 2. #% ( H2S leaving )\n", + "Pstp = 101.325; #kPa\n", + "Tstp = 273.15; #K\n", + "Vstp = 22.414; #m**3/kmol\n", + "\n", + "# Calculation \n", + "Vstp1 = V * P * Tstp / ( T * Pstp)\n", + "N = Vstp1 / Vstp;\n", + "N1 = N * per1 / 100;\n", + "N2 = N - N1; # ( 2 = inerts)\n", + "Nleaving = N2 / ( 1 - per2 / 100);\n", + "N1leaving = per2 * Nleaving / 100;\n", + "mabsorbed = (N1 - N1leaving) * 34.08; #( molecular wt. = 34.08)\n", + "mgiven = 100. #kg/h\n", + "Vactual = mgiven * V / mabsorbed;\n", + "Nactual = Nleaving * Vactual / V; # actual moles leaving\n", + "Vstpl = Nactual * Vstp; # volume leaving at STP\n", + "P2 = 500. #kPa\n", + "T2 = 290. #K\n", + "V2 = Vstpl * Pstp * T2 / ( P2 * Tstp);\n", + "Precovery = (N1 - N1leaving)*100 / N1;\n", + "\n", + "# Result\n", + "print \"(a)Volume of gas entering per hour %.2f m**3/h\"%Vactual\n", + "print \"(b)Volume of gas leaving per hour %.2f m**3/h\"%V2\n", + "print \"(c)Percentage recovery of H2S %.2f %%\"%Precovery\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Volume of gas entering per hour 68.63 m**3/h\n", + "(b)Volume of gas leaving per hour 62.89 m**3/h\n", + "(c)Percentage recovery of H2S 91.84 %\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 pageno : 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "#N2 + 3H2 = 2NH3\n", + "V1 = 100. #m**3 ( 1 = N2) cubic metres\n", + "V2 = V1 * 3 # ( According to Avagadros principle, equal\n", + " #volumes of all gases under similar condition\n", + " # contains same no. of moles)\n", + " \n", + "print \"(a)Volume of hydrogen required at same condition = \",V2,\"m**3\"\n", + "P1 = 20. #bar \n", + "T1 = 350. #K nitrogen\n", + "P2 = 5. #bar\n", + "T2 = 290. #K hydrogen\n", + "\n", + "# Calculation and Result\n", + "V3 = 3 * V1 * P1 * T2 / ( P2 * T1)\n", + "print \"(b)Volume required at 50 bar and 290K = %.3f m**3\"%V3\n", + "m = 1000 #kg ( ammonia )\n", + "N = m / 17.03 #kmol\n", + "N1 = N/2. # ( nitrogen)\n", + "N2 = N * 3 / 2. #(hydrogen)\n", + "P3 = 50. #bar\n", + "T3 = 600. #K\n", + "Pstp = 1.01325 #bar\n", + "Tstp = 273.15 #K\n", + "Vstp = 22.414 #m**3/kmol\n", + "V1stp = N1 * Vstp\n", + "V4 = V1stp * Pstp * T3 / (P3 * Tstp) # ( nitrogen at 50 bar and 600K)\n", + "V5 = V4 * 2 # ( ammonia at 50 bar and 600K)\n", + "V6 = V4 * 3 # ( hydrogen at 50 bar and 600K)\n", + "print \"(c)Volume of nitrogen at 50 bar and 600K = %.1f m**3\"%V4\n", + "print \" Volume of hydrogen at 50 bar and 600K = %.1f m**3\"%V6\n", + "print \" Volume of ammonia at 50 bar and 600K = %.1f m**3\"%V5\n", + "\n", + "\n", + "# note : answers may vary because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Volume of hydrogen required at same condition = 300.0 m**3\n", + "(b)Volume required at 50 bar and 290K = 994.286 m**3\n", + "(c)Volume of nitrogen at 50 bar and 600K = 29.3 m**3\n", + " Volume of hydrogen at 50 bar and 600K = 87.9 m**3\n", + " Volume of ammonia at 50 bar and 600K = 58.6 m**3\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12 pageno : 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "N = 100. #kmol producer gas\n", + "P1 = 25. #% ( Carbon monoxide )\n", + "P2 = 4. #% ( Carbon Dioxide )\n", + "P3 = 3. #% ( Oxygen )\n", + "P4 = 68. #% ( Nitrogen )\n", + "\n", + "# Calculation \n", + "N1 = N * P1/100 \n", + "N2 = N * P2/100 \n", + "N3 = N * P3/100 \n", + "N4 = N * P4/100 \n", + "NC = N1 + N2 \n", + "m = NC * 12 \n", + "Ngas = N / m #moles of gas for 1 kg of Carbon\n", + "Vstp = 22.4143 #m**3/kmol\n", + "Vstp1 = Vstp * Ngas \n", + "P = 1. #bar\n", + "T = 290. #k\n", + "Pstp = 1.01325 #bar\n", + "Tstp = 273.15 #K\n", + "V = T * Vstp1 * Pstp / (Tstp * P ) \n", + "print \"(a)Volume of gas at 1 bar and 290 K per kg Carbon = %.2f m**3\"%V\n", + "\n", + "#CO + 1/2 * O2 = CO2\n", + "Nrequired = N1/2 - N3 #(oxygen required)\n", + "Nsupplied = Nrequired * 1.2 \n", + "PO1 = 21. #% ( Oxygen percent in air)\n", + "Nair = Nsupplied * 100/PO1 \n", + "V1 = 100. #m**3 \n", + "Vair = V1 * Nair / N \n", + "print \"(b)Volume of air required = %.2f m**3\"%Vair\n", + "NCO2 = N2 + N1 \n", + "NO2 = Nsupplied - Nrequired \n", + "NN2 = N4 + (Vair * (1 - PO1/ 100)) \n", + "Ntotal = NCO2 + NO2 + NN2 \n", + "PCO2 = NCO2 * 100 / Ntotal \n", + "PO2 = NO2 * 100 / Ntotal \n", + "PN2 = NN2 * 100 / Ntotal \n", + "\n", + "# Result\n", + "print \"Percent composition of Carbon Dioxide = %.2f %%\"%PCO2\n", + "print \"Percent composition of Oxygen = %.2f %%\"%PO2\n", + "print \"Percent composition of Nitrogen = %.2f %%\"%PN2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Volume of gas at 1 bar and 290 K per kg Carbon = 6.93 m**3\n", + "(b)Volume of air required = 54.29 m**3\n", + "Percent composition of Carbon Dioxide = 20.45 %\n", + "Percent composition of Oxygen = 1.34 %\n", + "Percent composition of Nitrogen = 78.21 %\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13 pageno : 94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "#4HCl + O2 = 2Cl2 + 2H2O\n", + "n = 1. #mol ( Basis 1 mol of HCl )\n", + "NO2 = n / 4 \n", + "NO2supp = 1.5 * NO2 \n", + "Nair = NO2supp * 100 / 21 \n", + "V = 100. #m**3\n", + "Vair = V * Nair / n \n", + "print \"(a)Volume of air admitted = %.1f m**3\"%Vair\n", + "\n", + "# Calculation \n", + "P1 = 80. #% ( HCl converted)\n", + "Ncon = n * P1 /100 \n", + "N2 = Ncon/4 # oxygen required\n", + "NH2O = Ncon / 2 \n", + "NCl2 = Ncon / 2 \n", + "nHCl = n - Ncon \n", + "nO2 = NO2supp - N2 \n", + "Nnitro = Nair - NO2supp \n", + "Ntotal = nHCl + nO2 + NH2O + NCl2 + Nnitro \n", + "V1 = V * Ntotal \n", + "P1 = 1. #bar\n", + "T1 = 290. #K\n", + "P2 = 1.2 #bar\n", + "T2 = 400. #K\n", + "V2 = V1 * P1 * T2 / ( P2 * T1) \n", + "print \"(b)Volume of gas leaving = %.2f m**3\"%V2\n", + "VCl2 = NCl2 * V \n", + "Pstp = 1.01325 #bar\n", + "Tstp = 273. #K\n", + "Vstp = 22.4143 #m**3/kmol\n", + "Vstp1 = Tstp * P1 * VCl2 / (T1 * Pstp) \n", + "Nstp = Vstp1/Vstp \n", + "m = Nstp * 70.90 \n", + "print \"(c)Kilograms of Chlorine produced = %.1f kg\"%m\n", + "Ntotaldry = nHCl + nO2 + NCl2 + Nnitro #dry basis\n", + "p1 = nHCl*100/Ntotaldry \n", + "p2 = nO2*100/Ntotaldry \n", + "p3 = NCl2*100/Ntotaldry \n", + "p4 = Nnitro*100/Ntotaldry \n", + "\n", + "# Result\n", + "print \"(d)Percent composition of HCl in exit stream = %.2f %%\"%p1\n", + "print \" Percent composition of Oxygen in exit stream = %.1f %%\"%p2\n", + "print \" Percent composition of Chlorine in exit stream = %.2f %%\"%p3\n", + "print \" Percent composition of nitrogen in exit stream = %.2f %%\"%p4\n", + "\n", + "# note : answers may vary because of rouding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Volume of air admitted = 178.6 m**3\n", + "(b)Volume of gas leaving = 297.21 m**3\n", + "(c)Kilograms of Chlorine produced = 117.6 kg\n", + "(d)Percent composition of HCl in exit stream = 9.15 %\n", + " Percent composition of Oxygen in exit stream = 8.0 %\n", + " Percent composition of Chlorine in exit stream = 18.30 %\n", + " Percent composition of nitrogen in exit stream = 64.54 %\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.14 pageno: 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "# CO2 = CO + 1/2 * O2\n", + "P1 = 1. #bar \n", + "T1 = 3500. #K pressure\n", + "P2 = 1. #bar \n", + "T2 = 300. #K heated\n", + "V2 = 25. #L CO2\n", + "\n", + "# Calculation \n", + "V1 = V2 * P2 * T1 / ( P1 * T2 ) \n", + "\n", + "# Result\n", + "print \"(a)Final volume of gas if no dissociation occured = %.3f m**3\"%(V1/1000)\n", + "Pstp = 1.01325 #bar\n", + "Tstp = 273. #K\n", + "Vstp = 22.4143 #m**3\n", + "N2 = V2 * P2 * Tstp / ( Vstp * Pstp * T2) \n", + "\n", + "# let x be the fraction dissociated, then after dissociation,\n", + "# CO2 = (1 - x)mol, CO = xmol, O2 = (0.5*x)mol\n", + "#total moles = 1 - x + x + o.5 * x = 1 + 0.5 * x \n", + "V = 350. #L\n", + "N1 = V * P1 * Tstp / (Vstp * Pstp * T1) \n", + "\n", + "# 1 + 0.5 * x = N1, therefore\n", + "x = (N1 - 1) / 0.5 \n", + "p = x*100 \n", + "print \"(b)CO2 converted = %.f %%\"%p\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Final volume of gas if no dissociation occured = 0.292 m**3\n", + "(b)CO2 converted = 40 %\n" + ] + } + ], + "prompt_number": 28 + } + ], + "metadata": {} + } + ] +}
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