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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 2 : Water Systems"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.1 Page No : 2-5"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\n",
      "#initialisation of variables\n",
      "w = 3000.\t#sq ft\n",
      "w1 = 2000.\t#sq ft\n",
      "w2 = 1000.\t#sq ft\n",
      "r = 15. \t#in\n",
      "a = 12.\t    #in\n",
      "h = 7.5\t    #in\n",
      "\t\n",
      "#CALCULATIONS\n",
      "G = w*(r/a)*h\t#gal\n",
      "g = w1*(r/a)*h\t#gal\n",
      "g1 = w2*(r/a)*h\t#gal\n",
      "\t\n",
      "#RESULTS\n",
      "print 'the normally be stored to tide the supply over dry spells = %.0f gal'%(G)\n",
      "#wrong ans in book"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the normally be stored to tide the supply over dry spells = 28125 gal\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2 Page No : 2-9"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\n",
      "#initialisation of variables\n",
      "m = 17.378   \t#mg\n",
      "h = 20.  \t    #in/sq mile \n",
      "d = 365.\t    #in\n",
      "s = 0.75\t    #percent\n",
      "a = 100.\t    #sq miles\n",
      "p = 750000.\t    #gpd per sq mile\n",
      "t = 180.\t    #in\n",
      "c = 150.\t    #gpcd\n",
      "n = 64600. \t    #gpd per sq mile\n",
      "\t\n",
      "#CALCULATIONS\n",
      "R = h*m\t#mg\n",
      "A = R/d\t#mgd\n",
      "S = s*a*t\t#billion gal\n",
      "Q = a*p/c\t#gpd\n",
      "P = a*n/c\t#people against\n",
      "\t\n",
      "#RESULTS\n",
      "print 'the surface water sheds and storage requirements = %.0f mg'%(R)\n",
      "print 'the well watered sections of north america = %.1f billion gal'%(S/1000)\n",
      "print 'the average consumption populations = %.0f gpd'%(Q)\n",
      "print 'the presence of proper storage = %.0f people against'%(round(P,-3))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the surface water sheds and storage requirements = 348 mg\n",
        "the well watered sections of north america = 13.5 billion gal\n",
        "the average consumption populations = 500000 gpd\n",
        "the presence of proper storage = 43000 people against\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.3 Page No : 2-14"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\n",
      "#initialisation of variables\n",
      "w = 20  \t#ft\n",
      "r = 3\t    #ft a day\n",
      "g = 500\t    #ft\n",
      "g1 = 1000\t    #ft\n",
      "h = 7.5/1440\t#ft\n",
      "p = 7.5/1000000\t#ft\n",
      "r1 = 2\t        #ft a day\n",
      "\t\n",
      "#CALCULATIONS\n",
      "W1 = w*g1*r*h\t#gpm\n",
      "W2 = w*g1*r1*r*p\t#mgd\n",
      "\t\n",
      "#RESULTS\n",
      "#wrong ans in book\n",
      "print 'the ground water laterally  = %.2f gpm'%(W1)\n",
      "print 'the water from both sides = %.2f mgd'%(W2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the ground water laterally  = 312.50 gpm\n",
        "the water from both sides = 0.90 mgd\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.4 Page No : 2-18"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\n",
      "#initialisation of variables\n",
      "p1 = 10.\t    #mgd\n",
      "p2 = 6940.\t#gpm\n",
      "w = 67000.\t#people\n",
      "d = 2.\t    #min\n",
      "v = d*p2/d\t#gal\n",
      "v1 = 98.2\t#cu ft each\n",
      "q = 30.\t    #min\n",
      "q1 = q*p2/d\t#gal\n",
      "q2 = 13900.\t#cu ft\n",
      "a = 1390.\t#sq ft\n",
      "s = 2.\t    #hr\n",
      "s1 = 120*p2/d\t#gal\n",
      "s2 = 55700.\t#cu ft \n",
      "s3 = s2/8.\t#sq ft\n",
      "r = 3\t    #gpm/sq ft\n",
      "r1 = 6\t    #rapid\n",
      "\t\n",
      "#CALCULATIONS\n",
      "D = math.sqrt(v1*4/math.pi)\t#ft\n",
      "S = p2/s3\t#gpm/sq ft\n",
      "A = p2/(r1*r)\t#sq ft\n",
      "\t\n",
      "#RESULTS\n",
      "print 'the capacity of the components of a rapid sand filtration plant = %d sq ft'%(A)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the capacity of the components of a rapid sand filtration plant = 385 sq ft\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.5 Page No : 2-21"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\n",
      "#initialisation of variables\n",
      "r = 10000.\t#ft\n",
      "l = 400000\t#people\n",
      "q = 1000000.\t#mgd\n",
      "w = 100\t#gpcd\n",
      "w1 = 150\t#gpcd\n",
      "m = 50\t#percent\n",
      "g = 1.5\t#ft\n",
      "h1 = 2.32\t#cfs\n",
      "h2 = 139\t#cfs\n",
      "d = 12\t#ft\n",
      "c = 100\t#ft\n",
      "l = 10.8\t#ft\n",
      "l2 = 0.85\t#ft\n",
      "l1 = 1000\t#ft\n",
      "\t\n",
      "#CALCULATIONS\n",
      "a = r*w/q\t#mgd\n",
      "b = l*w1/q\t#mgd\n",
      "a1 = a*g\t#mgd\n",
      "b1 = b*g\t#mgd\n",
      "D = d*math.sqrt(h1/math.pi)\t#in\n",
      "D1 = d*math.sqrt(h2/math.pi)\t#in\n",
      "L = l/l1\t#ft\n",
      "L1 = l2/l1\t#ft\n",
      "\t\n",
      "#RESULTS\n",
      "print 'the higher loss of head in small conduits at equal velocity = %f ft'%(L1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the higher loss of head in small conduits at equal velocity = 0.000850 ft\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.6 Page No : 2-25"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\n",
      "#initialisation of variables\n",
      "a = 12\t#in\n",
      "b = 24\t#in\n",
      "r = 500\t#gpm\n",
      "d = 200.\t#gpcd\n",
      "d1 = 150.\t#gpcd\n",
      "p1 = 113\t#sq in\n",
      "p2 = 425\t#sq in\n",
      "v1 = 3\t    #fps\n",
      "v2 = 2.35\t#cfs\n",
      "v3 = 9.42\t#cfs\n",
      "h = 646000\t#gpd\n",
      "w = 720000\t#gpd\n",
      "\t\n",
      "#CALCULATIONS\n",
      "D1 = v2*h\t#gpd\n",
      "D2 = v3*h\t#gpd\n",
      "W1 = D1-w\t#gpd\n",
      "W2 = D2-w\t#gpd\n",
      "R1 = D1/d\t#people\n",
      "R2 = D2/d\t#people\n",
      "S = W1/d1\t#people\n",
      "S1 = W2/d1\t#people\n",
      "\t\n",
      "#RESULTS\n",
      "print 'the absence of fire service for a maximum draft = %d gpd'%(round(D2,-3))\n",
      "print 'The residential fire flow requirements = %.0f gpd'%(round(W2,-3))\n",
      "\n",
      "# note: answers are correct.  please check using calculator."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the absence of fire service for a maximum draft = 6085000 gpd\n",
        "The residential fire flow requirements = 5365000 gpd\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.7 Page No : 2-28"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\n",
      "#initialisation of variables\n",
      "w = 100000.\t#ft\n",
      "c = 250.  \t#per capita\n",
      "p1 = 0.3\t#percent\n",
      "p2 = 0.1\t#percent\n",
      "p3 = 0.60\t#percent\n",
      "w1 = 15.\t#mgd\n",
      "\t\n",
      "#CALCULATIONS\n",
      "T = c*w\t#$\n",
      "W = p1*T\t#$\n",
      "W1 = p2*T\t#$\n",
      "W2 = p3*T\t#$\n",
      "W3 = ((w1)**(2./3))*(T/100)\t#$\n",
      "\n",
      "#RESULTS\n",
      "print 'the replacement cost of the water of a city = %.0f $'%(round(W3,-5))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the replacement cost of the water of a city = 1500000 $\n"
       ]
      }
     ],
     "prompt_number": 23
    }
   ],
   "metadata": {}
  }
 ]
}