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|
{
"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": {}
}
]
}
|
