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diff --git a/Water_and_Wastewater_Engineering/ch3.ipynb b/Water_and_Wastewater_Engineering/ch3.ipynb new file mode 100644 index 00000000..1294bb03 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch3.ipynb @@ -0,0 +1,348 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0ac87828ad067f0685215c3ea3f783dc1f30af884f5edc144a109aec1b63e4e7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : Wastewater Systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page No : 3-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "v = 2.5\t#fps\n", + "c = 0.013\t#gpd\n", + "p = 300\t#gpd\n", + "d = 50\t#per care\n", + "m = 5.20\t#ft\n", + "a = 1000\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "C = (math.pi*64)/(4*144)*v*646000\t#gpd\n", + "M = m/a\t#ft\n", + "P = C/p\n", + "A = P/d\t#acres\n", + "\t\n", + "#RESULTS\n", + "print 'the acres will it drain if the population density = %.1f acres'%(A)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the acres will it drain if the population density = 37.6 acres\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 3-14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "a = 37.4\t#acres\n", + "r = 2.\t#in\n", + "p = 30.\t#min\n", + "v = 3\t#fps\n", + "r1 = 0.6\t#in\n", + "h = 1.0\t#cfs\n", + "p1 = 50\t#percent\n", + "q = 646000\t#gpd\n", + "\t\n", + "#CALCULATIONS\n", + "R = r*r1*a\t#cfs\n", + "A = R/v\t#sq ft\n", + "D = 12*math.sqrt(4*A/math.pi)\t#in\n", + "P = r*r1*q/p1\t#gpcd\n", + "\t\n", + "#RESULTS\n", + "print 'the per capita storm runoff for a population density = %.0f gpcd'%(round(P,-1))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the per capita storm runoff for a population density = 15500 gpcd\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No : 3-16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "w = 1.0\t#cfs\n", + "w1 = 3.0\t#cfs\n", + "w2 = 45.0\t#cfs\n", + "v = 3.0\t#fps\n", + "h = 144\t#ft\n", + "D = 12*math.sqrt(4*w/(math.pi*w1))\t#in\n", + "d1 = 1.95\t#cfs\n", + "D1 = 12*math.sqrt(4*d1)/(math.pi*v)\t#in\n", + "d2 = 41.6\t#cfs\n", + "D2 = 12*math.sqrt(4*d2)/(math.pi*w1)\t#ins\n", + "\t\n", + "#CALCULATIONS\n", + "C = math.pi*(D)**2*3/(4*h)\t#cfs\n", + "C1 = math.pi*(1./4)*3\t#cfs\n", + "V = (d2*4)/(math.pi*4**2)\t#fps\n", + "\t\n", + "#RESULTS\n", + "print 'The minimum dry-weather flow = %.2f cfs'%(C)\n", + "print 'The maximum dry-weather flow in excess actual capacity = %.1f cfs'%(C1)\n", + "print 'the storm flow in axcess of maximum dry-weather flow = %.1f fps'%(V)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum dry-weather flow = 1.00 cfs\n", + "The maximum dry-weather flow in excess actual capacity = 2.4 cfs\n", + "the storm flow in axcess of maximum dry-weather flow = 3.3 fps\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 3-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "t = 0.8\t#mgd\n", + "d = 8000.\t#people\n", + "a = 2.\t#hr\n", + "v = 800000.\t#ft\n", + "h = 10.\t#ft\n", + "a1 = 4.\t#in\n", + "d1 = 1.\t#sq ft per capita\n", + "a3 = 3.\t#mgad\n", + "\t\n", + "#CALCULATIONS\n", + "V = v*(a/24)/a\t#gal\n", + "S = V/h\t#sq ft\n", + "S1 = (v/h)/S\t#gpd per sq ft\n", + "V1 = a*d/h\t#cu ft\n", + "D = d/S\t#ft\n", + "E = d1*d/a1\t#sq ft\n", + "A = t/a3\t#acre\n", + "\t\n", + "#RESULTS\n", + "print 'the capacity of the components of a small trickling-filter = %.2f acre'%(A)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the capacity of the components of a small trickling-filter = 0.27 acre\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No : 3-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "w = 2000.\t#sq miles\n", + "r = 0.1 \t#cfs\n", + "d = 80000.\t#ft\n", + "p = 100.\t#gpd\n", + "p1 = 80.\t#ft\n", + "p2 = 340.\t#percent\n", + "h = 646000.\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "L = r*w\t#cfs\n", + "R = 6*p1/1.4\t#cfs\n", + "P = p1*(p2-L)/p2\t#percent\n", + "D = (d*p)\t#gpd\n", + "D1 = (L*h)\t#gpd\n", + "\t\n", + "#RESULTS\n", + "print \"Low-water flow = %d cfs\"%(0.1*w)\n", + "print \"Required flow for disposal of domestic sewage if it i left untreated = %d cfs\"%(round(6*80/1.4,-1))\n", + "print 'the percent of removal of pollutional load needed = %.0f percent'%(P)\n", + "print \"Dilution ratio = %d : %d\"%((d*p)/(d*p),(200*h)/(d*p))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Low-water flow = 200 cfs\n", + "Required flow for disposal of domestic sewage if it i left untreated = 340 cfs\n", + "the percent of removal of pollutional load needed = 33 percent\n", + "Dilution ratio = 1 : 16\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page No : 3-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p = 10000.\t#people\n", + "p1 = 4. \t#ft\n", + "w = 10. \t#in\n", + "s = 80. \t#gpcd\n", + "h = 43560.\t#ft\n", + "p2 = 20. \t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "R = ((w/12)*7.5*h)/365\t#gpad\n", + "A = p*s/R\t#acres\n", + "A1 = 1.7\t#sq miles\n", + "P = p/500\t#acres\n", + "D = p2*h*4*7.48/(p*s)\t#days\n", + "\t\n", + "#RESULTS\n", + "print 'the detention period in ponds = %.0f days'%(D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the detention period in ponds = 33 days\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7 Page No : 3-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p = 100000.\t#people\n", + "a = 75. \t#$\n", + "a2 = 47. \t#in\n", + "b = 10. \t#in\n", + "\t\n", + "#CALCULATIONS\n", + "P = a*p\t#people\n", + "S = ((a2)*(b**5))/(b)**(1./4)\t#$\n", + "\t\n", + "#RESULTS\n", + "print 'the money is inversed in the sanitary sewerage system = %.0f $'%(round(S,-5))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the money is inversed in the sanitary sewerage system = 2600000 $\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
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