{ "metadata": { "name": "", "signature": "sha256:209dfa66c415d9b6d77e6ef0fd2310df1d6593c2b60b5941934cee9e5b7d9fa6" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 15 : Wastewater Collection" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.1 Page No : 15-4" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\t\n", "#initialisation of variables\n", "q = 0.25\t#in\n", "Q = 0.34\t#in\n", "r = 0.76\t#in\n", "v = 0.83\t#in\n", "n = 0.78\t#in\n", "r1 = 0.84\t#in\n", "v1 = 0.70\t#in\n", "w = 2\t#in\n", "q1 = 0.056\t#in\n", "d = 0.16\t#in\n", "v2 = 0.53\t#in\n", "n1 = 0.80\t#in\n", "d1 = 0.18\t#in\n", "n2 = 0.46\t#in\n", "\t\n", "#CALCULATIONS\n", "V = v*w\t#fps\n", "N = v1*w\t#fps\n", "V1 = v2*w\t#fps\n", "V2 = n2*w\t#fps\n", "\t\n", "#RESULTS\n", "print 'The one fourth their full flow = %.2f fps'%(N)\n", "print 'The one enghteenth their full flow = %.2f fps'%(V2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The one fourth their full flow = 1.40 fps\n", "The one enghteenth their full flow = 0.92 fps\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.2 Page No : 15-9" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\t\n", "#initialisation of variables\n", "v = 2.5\t#fps\n", "N = 0.015\t#fps\n", "a = (40+27)\t#in\n", "b = (40*27+27*19)/a\n", "c = 0.440\t#cfs\n", "w = 49*0.09/100\t#cfs\n", "g = 0.008\t#percent\n", "Q = 0.82\t#cfs\n", "r = 0.795\t#cfs\n", "t = 2.35*1.16\t#fps\n", "d1 = 113.20-113.03\t#ft\n", "d2 = 12\t#ft\n", "\t\n", "#CALCULATIONS\n", "R = r/Q\t#cfs\n", "D = g*r\t#in\n", "D2 = d1*d2\t#in\n", "\t\n", "#RESULTS\n", "print 'The required capacity and find the slope size and hydraulic characteristics of the system = %.0f in'%(D2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The required capacity and find the slope size and hydraulic characteristics of the system = 2 in\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.3 Page No : 15-17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\t\n", "#initialisation of variables\n", "p = 20\t#min\n", "N = 0.012\t#in\n", "k = 2.19\t#min\n", "l = k+1.97\t#min\n", "q = 340/(60*3.94)\t#min\n", "r = 2.56*0.508\t#min\n", "del1 = 0.42\t#min\n", "j = 84.28\t#min\n", "w1 = 0.92\t#min\n", "\t\n", "#CALCULATIONS\n", "r1 = r*k\t#cfs\n", "w = p+q\t#min\n", "G = j-del1\t#min\n", "S = (G-w1)\t#min\n", "\t\n", "#RESULTS\n", "print 'The required capacity and find the slop size and hydraulic = %.2f min'%(S)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The required capacity and find the slop size and hydraulic = 82.94 min\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.4 Page No : 15-25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\t\n", "#initialisation of variables\n", "a = 42.\t#in\n", "d = 45.\t#mgd\n", "d1 = 0.75\t#in\n", "s = 60.\t#ft\n", "p1 = 9.\t#in\n", "p2 = 8.4\t#in\n", "p3 = 9.\t#in\n", "c1 = 13*63.6\t#sq in\n", "c2 = 9*55.4\t#sq in\n", "c3 = 9.21\t#sq ft\n", "M = d*1.547\t#cfs\n", "v = M/c3\t#fps\n", "g = 0.025*32.2\t#ft/sec**2\n", "\t\n", "#CALCULATIONS\n", "F = v/math.sqrt(g*(p1/12))\t#ft\n", "S = s/d1\t#in\n", "\t\n", "#RESULTS\n", "print 'the port near the end of the diffuser pipe = %.2f in'%(F)\n", "\n", "# note : rounding off error." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the port near the end of the diffuser pipe = 9.73 in\n" ] } ], "prompt_number": 2 } ], "metadata": {} } ] }