{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 14: Water Treatment"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 1, Page No:378"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"wt_CaSO4 = 160 # mg / l\n",
"\n",
"# Solution\n",
"hardness = 100 * wt_CaSO4 / 136.\n",
"print \"The hardness is\", \"{:.2f}\".format(hardness), \"mg / L of CaCO3 eqv.\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The hardness is 117.65 mg / L of CaCO3 eqv.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 2, Page No:378"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"wt1 = 9.3 # mg / L\n",
"wt2 = 17.4 # mg / L\n",
"wt3 = 8.7 # mg / L\n",
"wt4 = 12.6 # mg / L\n",
"\n",
"# Solution\n",
"temp_hardness = wt1 * 100 / 146. + wt2 * 100 / 162.\n",
"per_hardness = wt3 * 100 / 95. + wt4 * 100 / 136.\n",
"total_hardness = temp_hardness + per_hardness\n",
"\n",
"print \"Temporary hardness:\", \"{:.2f}\".format(temp_hardness), \"mg / L\"\n",
"print \"Total hardness:\", \"{:.2f}\".format(total_hardness), \"mg / L\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Temporary hardness: 17.11 mg / L\n",
"Total hardness: 35.53 mg / L\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 3, Page No:378"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"wt1 = 32.4 # mg / L\n",
"wt2 = 29.2 # mg / L\n",
"wt3 = 13.5 # mg / L\n",
"\n",
"# Solution\n",
"temp_hardness = wt1 * 100 / 162. + wt2 * 100 / 146.\n",
"per_hardness = wt3 * 100 / 136.\n",
"\n",
"print \"Temporary hardness:\", \"{:.2f}\".format(temp_hardness), \"mg / L\"\n",
"print \"Permanent hardness:\", \"{:.2f}\".format(per_hardness), \"mg / L\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Temporary hardness: 40.00 mg / L\n",
"Permanent hardness: 9.93 mg / L\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 4, Page No:379"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"i1 = 180 # mg / L, CaCl2\n",
"i2 = 210 # mg / L, Ca(NO3)2\n",
"i3 = 123 # mg / L, MgSO4\n",
"i4 = 90 # mg / L, Mg(HCO3)2\n",
"\n",
"# Solution\n",
"i1_req = i1 * 100 / 111.\n",
"i2_req = i2 * 100 / 164.\n",
"i3_req = i3 * 100 / 120.\n",
"i4_req = i4 * 100 / 146.\n",
"\n",
"lime_req = 74 / 100. * (2 * i4_req + i3_req) * 100 / 70. * 10000\n",
"soda_req = 106 / 100. * (i1_req + i3_req + i2_req) * 100 / 80. * 10000\n",
"\n",
"print \"Lime Required\", \"{:.1e}\".format(lime_req), \"mg\",\n",
"print \"=\", \"{:.1f}\".format(lime_req / 10 ** 6), \"kg\"\n",
"print \"Soda Required\", \"{:.1e}\".format(soda_req), \"mg\",\n",
"print \"=\", \"{:.1f}\".format(soda_req / 10 ** 6), \"kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Lime Required 2.4e+06 mg = 2.4 kg\n",
"Soda Required 5.2e+06 mg = 5.2 kg\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 5, Page No:379"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"wt1 = 32.4 # mg / L, Ca(HCO3)2\n",
"wt2 = 29.29 # mg / L, Mg(HCO3)2\n",
"wt3 = 13.5 # mg / L, CaSO4\n",
"\n",
"# Solution\n",
"wt1_equi = wt1 * 100 / 162.\n",
"wt2_equi = wt2 * 100 / 146.\n",
"wt3_equi = wt3 * 100 / 136.\n",
"\n",
"temp_hardness = wt1_equi + wt2_equi\n",
"perm_hardness = wt3_equi\n",
"\n",
"print \"Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is\",\n",
"print int(temp_hardness), \"ppm\"\n",
"print \"Permanent hardness [due to CaSO4] is\", \"{:.1f}\".format(perm_hardness), \"ppm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is 40 ppm\n",
"Permanent hardness [due to CaSO4] is 9.9 ppm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 6, Page No:380"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"v1 = 150 # litres, NaCl\n",
"\n",
"# Solution\n",
"v_hardwater = 22500 * v1 / 3 / 0.6 / 58.5\n",
"\n",
"print \"The amount of hard water that can be softened using this softner is\",\n",
"print int(v_hardwater), \"litres\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The amount of hard water that can be softened using this softner is 32051 litres\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 7, Page No:380"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"v1 = 30 # litres, NaCl\n",
"w = 1500 # mg / L, NaCl\n",
"\n",
"# Solution\n",
"hardness = 45 * 50 / 58.5 * 1000 / 1000\n",
"print \"Hardness of water is\", \"{:.2f}\".format(hardness), \"ppm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Hardness of water is 38.46 ppm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 8, Page No:381"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"v1_water = 50 # ml, water\n",
"w1_CaCO3 = 1.5 # mg, pure CaCO3\n",
"v1_EDTA = 44 # ml, EDTA\n",
"v2_EDTA = 40 # ml, EDTA\n",
"v2_water = 20 # ml, water\n",
"\n",
"# Solution\n",
"EDTA_1ml = v1_water * w1_CaCO3 / v1_EDTA\n",
"hardwater_40ml = v2_water * 1.704\n",
"total_hardness0 = hardwater_40ml * 1000 / 40\n",
"total_hardness1 = total_hardness0 * 0.07\n",
"\n",
"print \"Total hardness is\", \"{:.2f}\".format(total_hardness1), \"\u00b0Cl\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total hardness is 59.64 \u00b0Cl\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 9, Page No:381"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constants\n",
"Fe = 56\n",
"S = 32\n",
"O = 16\n",
"Ca = 40\n",
"C = 12\n",
"\n",
"# Solution\n",
"hardness100 = Fe + S + O * 4\n",
"\n",
"print \"215 ppm of hardness is\", \"{:.1f}\".format(hardness100 * 215 / 100.),\n",
"print \"ppm of FeSO4\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"215 ppm of hardness is 326.8 ppm of FeSO4\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 10, Page No:381"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"v1 = 50. # ml, hardwater\n",
"v2 = 15 # ml, EDTA\n",
"m = 0.01 # M, EDTA\n",
"\n",
"# Solution\n",
"M = v2 * m / v1\n",
"N = M * 2\n",
"S = N * 50 * 1000\n",
"\n",
"print \"Molarity of hardness is\", M, \"M\"\n",
"print \"Normality of hardness is\", N, \"N\"\n",
"print \"Strength of hardness is\", S, \"ppm or mg / L\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Molarity of hardness is 0.003 M\n",
"Normality of hardness is 0.006 N\n",
"Strength of hardness is 300.0 ppm or mg / L\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 11, Page No:382"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variable\n",
"C = 16.5 # ppm, CO3-2\n",
"\n",
"# Solution\n",
"Molarity = C * 10 ** - 6 / 60.\n",
"\n",
"print \"Molarity of CO3-2 is\", \"{:.1e}\".format(Molarity), \"mol / L\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Molarity of CO3-2 is 2.7e-07 mol / L\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}