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diff --git a/sample_notebooks/ajinkyakhair/chapter17.ipynb b/sample_notebooks/ajinkyakhair/chapter17.ipynb new file mode 100755 index 00000000..2e81c43d --- /dev/null +++ b/sample_notebooks/ajinkyakhair/chapter17.ipynb @@ -0,0 +1,483 @@ +{ + "metadata": { + "celltoolbar": "Raw Cell Format", + "name": "", + "signature": "sha256:8ade9358728d8873e3787f09037539bd355e8453094fed25ae256529fe52d8d4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 17: Water Treatment" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1,Page number 369" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "m1 = 146 #mass of Mg(HCO3)2\n", + "\n", + "m2 = 162 #mass of Ca(HCO3)2\n", + "\n", + "m3 = 95 #mass of MgCl2\n", + "\n", + "m4 = 136 #mass of CaSO4\n", + "\n", + "amnt_1 = 7.5 #amount of Mg(HCO3)2 in mg/l\n", + "\n", + "amnt_2 = 16 #amount of Ca(HCO3)2 in mg/l\n", + "\n", + "amnt_3 = 9 #amount of MgCl2 in mg/l\n", + "\n", + "amnt_4 = 13.6 #amount of CaSO4 in mg/l\n", + "\n", + "temp_hard= (amnt_1*100./m1)+(amnt_2*100./m2)\n", + "\n", + "perm_hard= (amnt_3*100./m3)+(amnt_4*100./m4)\n", + "\n", + "total= temp_hard +perm_hard\n", + "\n", + "print\"the temporary hardness is =\",temp_hard,\"mg/l\"\n", + "\n", + "print\"the total hardness is =\",total,\"mg/l\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the temporary hardness is = 15.0135295112 mg/l\n", + "the total hardness is = 34.4872137218 mg/l\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 2,Page number 369" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "m1= 136 # mass of FeSO4\n", + "\n", + "m2 = 100 #mass of CaCO3\n", + "\n", + "#for 100 ppm hardness FeSO4 required per 10**6 parts of water is 136 parts\n", + "#for 200 ppm hardness\n", + "\n", + "amt= m1*200./m2\n", + "\n", + "print\"the amount of FeSO4 required is =\",amt,\"mg/l\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the amount of FeSO4 required is = 272 mg/l\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 3,Page number 369" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "conc = 15.6 *10**-6 #concentration of (CO3)-2\n", + "\n", + "m = 60 #mass of CO3\n", + "\n", + "Molarity= conc*100./m\n", + "\n", + "print\"the molarity of (CO3)-2 is =\",Molarity,\"M\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the molarity of (CO3)-2 is = 2.6e-05 M\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 4,Page number 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "m1 = 146 #mass of Mg(HCO3)2\n", + "\n", + "m2 = 162 #mass of Ca(HCO3)2\n", + "\n", + "m3 = 111 #mass of CaCl2\n", + "\n", + "m4 = 120 #mass of MgSO4\n", + "\n", + "m5 = 136 #mass of CaSO4\n", + "\n", + "amnt_1 = 12.5 #amount of Mg(HCO3)2 in ppm\n", + "\n", + "amnt_2 = 10.5 #amount of Ca(HCO3)2 in ppm\n", + "\n", + "amnt_3 = 8.2 #amount of CaCl2 in ppm \n", + "\n", + "amnt_4 = 2.6 #amount of MgSO4 in ppm\n", + "\n", + "amnt_5 = 7.5 #amount of CaSO4 in ppm\n", + "\n", + "temp_hard= (amnt_1*100./m1)+(amnt_2*100./m2)\n", + "\n", + "perm_hard= (amnt_3*100./m3)+(amnt_4*100./m4)+(amnt_5*100./m5)\n", + "\n", + "total= temp_hard +perm_hard\n", + "\n", + "print\"the temporary hardness is =\",temp_hard,\"mg/l\"\n", + "\n", + "print\"the permanent hardness is =\",perm_hard,\"mg/l\"\n", + "\n", + "print\"the total hardness is =\",total,\"mg/l\"\n", + "\n", + "v= 100 #volume of sample\n", + "\n", + "v_EDTA = total*v/1000 #volume of EDTA \n", + "\n", + "print\"the volume of M/100 EDTA required is =\",v_EDTA,\"ml\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the temporary hardness is = 15.0431253171 mg/l\n", + " the permanent hardness is = 15.0687599364 mg/l\n", + " the total hardness is = 30.1118852535 mg/l\n", + " the volume of M/100 EDTA required is = 3.01118852535 ml\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 5,Page number 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "v= 50000 #volume of water\n", + "\n", + "m1 = 84 #mass of MgCO3\n", + "\n", + "m2 = 100 #mass of CaCO3\n", + "\n", + "m3 = 95 #mass of MgCl2\n", + "\n", + "m4 = 111 #mass of CaCl2\n", + "\n", + "amnt_1 = 144 #amount of MgCO3 in ppm\n", + "\n", + "amnt_2 = 25 #amount of CaCO3 in ppm\n", + "\n", + "amnt_3 = 95 #amount of MgCl2 in ppm \n", + "\n", + "amnt_4 = 111 #amount of CaCl2 in ppm\n", + "\n", + "lime = (74./100)*(2*(amnt_1*100./m1)+(amnt_2*100./m2)+(amnt_3*100./m3))*v\n", + "\n", + "print\"the lime required is =\",lime,\"mg\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the lime required is = 17310714.2857 mg\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 6,Page number 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "v= 10**6 #volume of water\n", + "\n", + "m1 = 40.0 #mass of Ca+2\n", + "\n", + "m2 = 24.0 #mass of Mg+2\n", + "\n", + "m3 = 44.0 #mass of CO2\n", + "\n", + "m4 = 122.0 #mass of HCO3-\n", + "\n", + "amnt_1 = 20.0 #amount of Ca+2 in ppm\n", + "\n", + "amnt_2 = 25.0 #amount of Mg+2 in ppm\n", + "\n", + "amnt_3 = 30.0 #amount of CO2 in ppm \n", + "\n", + "amnt_4 = 150.0 #amount of HCO3- in ppm\n", + "\n", + "lime_1 = (74./100)*((amnt_2*100./m2)+(amnt_3*100./m3)+(amnt_4*100./m4))*v\n", + "\n", + "soda = (106./100)*((amnt_1*100./m1)+(amnt_2*100./m2)-(amnt_4*100./m4))*v\n", + "\n", + "print\"the lime required is =\",lime_1,\"mg\"\n", + "\n", + "print\"the soda required is =\",soda,\"mg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the lime required is = 218521485.345 mg\n", + "the soda required is = 33088797.8142 mg\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7,Page number 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "v= 150 #volume of NaCl \n", + "\n", + "conc = 150. #concentration of NaCl\n", + "\n", + "amnt =v*conc *100./117 #amnt of NaCl\n", + "\n", + "hard = 600. #hardness of water\n", + "\n", + "vol= amnt*1000./hard\n", + "\n", + "print\"the volume of water is =\",round(vol,4),\"litres\"\n", + "\n", + "#calculation mistake in textbook\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the volume of water is = 32051.2821 litres\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 8,Page number 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "strength = 10*0.85/9 #strength of EDTA\n", + "\n", + "#1000 ml EDTA solution == 1 g CaCO3\n", + "\n", + "#for 20 ml EDTA solution\n", + "\n", + "amnt= 20.*strength/1000\n", + "\n", + "#50 ml smple of water contains amnt CaCO3\n", + "\n", + "hard= amnt*10**6/50 #hardness of water \n", + "\n", + "print\"the hardness of water is =\",round(hard,3),\"ppm\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the hardness of water is = 377.778 ppm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 9,Page number 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Given Data\n", + "\n", + "m1 = 146 #mass of Mg(HCO3)2\n", + "\n", + "m2 = 162 #mass of Ca(HCO3)2\n", + "\n", + "m3 = 111 #mass of CaCl2\n", + "\n", + "m4 = 120 #mass of MgSO4\n", + "\n", + "m5 = 136 #mass of CaSO4\n", + "\n", + "amnt_1 = 12.5 #amount of Mg(HCO3)2 in ppm\n", + "\n", + "amnt_2 = 10.5 #amount of Ca(HCO3)2 in ppm\n", + "\n", + "amnt_3 = 8.2 #amount of CaCl2 in ppm \n", + "\n", + "amnt_4 = 2.6 #amount of MgSO4 in ppm\n", + "\n", + "amnt_5 = 7.5 #amount of CaSO4 in ppm\n", + "\n", + "temp_hard= ((amnt_1*100./m1)+(amnt_2*100./m2))*0.1\n", + "\n", + "perm_hard= ((amnt_3*100./m3)+(amnt_4*100./m4)+(amnt_5*100./m5))*0.1\n", + "\n", + "print\"the temporary hardness is =\",round(temp_hard,3),\"degree Fr\"\n", + "\n", + "print\"the permanent hardness is =\",round(perm_hard,3),\"degree Fr\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the temporary hardness is = 1.504 degree Fr\n", + "the permanent hardness is = 1.507 degree Fr\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
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