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{
"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": {}
}
]
}
|
