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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 15: Environmental Pollution and Control"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 1, Page No: 401"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constant\n",
"MM = 294 # Molar mass, K2Cr2O7\n",
"\n",
"# Variables\n",
"v_eff = 25 # cm ^ 3, effluent\n",
"v = 8.3 # cm ^ 3, K2Cr2O7\n",
"M = 0.001 # M, K2Cr2O7\n",
"\n",
"# Solution\n",
"w_O = v * 8 * 6 * M / 1000.\n",
"\n",
"print \"8.3 cm^3 of 0.006 N K2Cr2O7 =\", \"{:.3e}\".format(w_O), \"g of O2\"\n",
"print \"25 ml of the effluent requires\", \"{:.3e}\".format(w_O), \"g of O2\"\n",
"\n",
"cod = w_O * 10 ** 6 / 25.\n",
"print \"1 l of the effluent requires\", \"{:.2f}\".format(cod), \"g of O2\"\n",
"print \"COD of the effluent sample is\", \"{:.2f}\".format(cod), \"ppm or mg / L\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"8.3 cm^3 of 0.006 N K2Cr2O7 = 3.984e-04 g of O2\n",
"25 ml of the effluent requires 3.984e-04 g of O2\n",
"1 l of the effluent requires 15.94 g of O2\n",
"COD of the effluent sample is 15.94 ppm or mg / L\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 2, Page No: 401"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"v0 = 30 # cm^3, effluent\n",
"v1 = 9.8 # cm^3, K2Cr2O7\n",
"M = 0.001 # M, K2Cr2O7\n",
"\n",
"# Solution\n",
"O_30eff = 6 * 8 * v1 * M\n",
"print \"So 30 cm^3 of effluent contains =\", \"{:.4f}\".format(O_30eff), \"mg of O2\"\n",
"\n",
"cod = O_30eff * 1000 / 30.\n",
"\n",
"print \"1 l of the effluent requires\", cod, \"mg of O2\"\n",
"print \"COD of the effluent sample =\", cod, \"ppm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"So 30 cm^3 of effluent contains = 0.4704 mg of O2\n",
"1 l of the effluent requires 15.68 mg of O2\n",
"COD of the effluent sample = 15.68 ppm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 3, Page No: 401"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"v0 = 25 # ml, sewage\n",
"d0_O = 410 # ppm, dissolved oxygen\n",
"d1_O = 120 # ppm, dissolved oxygen\n",
"v1 = 50 # ml, sewage\n",
"\n",
"# Solution\n",
"print \"BOD = (DOb - DOi) * Dilution Factor\"\n",
"print \"BOD = (DOb - DOi) *\",\n",
"print \"(ml of sample after dilution) / (ml of sample before dilution)\"\n",
"\n",
"BOD = (d0_O - d1_O) * (v1 / v0)\n",
"print \"BOD =\", BOD, \"ppm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"BOD = (DOb - DOi) * Dilution Factor\n",
"BOD = (DOb - DOi) * (ml of sample after dilution) / (ml of sample before dilution)\n",
"BOD = 580 ppm\n"
]
}
],
"prompt_number": 10
}
],
"metadata": {}
}
]
}
|
