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| author | tslee | 2014-11-27 17:17:59 +0530 |
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| committer | tslee | 2014-11-27 17:17:59 +0530 |
| commit | 6e3407ba85ae84e1cee1ae0c972fd32c5504d827 (patch) | |
| tree | b89808101c39b1db1e3793eada2c8b702f856606 /Water_and_Wastewater_Engineering | |
| parent | 36a03d6d76bac315dba73b2ba9555c7e3fe0234f (diff) | |
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diff --git a/Water_and_Wastewater_Engineering/README.txt b/Water_and_Wastewater_Engineering/README.txt new file mode 100644 index 00000000..64c521dc --- /dev/null +++ b/Water_and_Wastewater_Engineering/README.txt @@ -0,0 +1,10 @@ +Contributed By: Sailesh Baria +Course: mca +College/Institute/Organization: Smart-Apps Solution Pvt Ltd +Department/Designation: Developer +Book Title: Water and Wastewater Engineering +Author: G. M. Fair, J. C. Geyer and D. A. Okun +Publisher: John Wiley, NewYork +Year of publication: 1966 +Isbn: 9780471251309 +Edition: 11
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch10.ipynb b/Water_and_Wastewater_Engineering/ch10.ipynb new file mode 100644 index 00000000..f78608b7 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch10.ipynb @@ -0,0 +1,111 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a8b74fc9d03d0c351dd92e0bc3a167aee10debeb1676b07b6cf10a252a49f523" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Groundwater Collection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1 Page No : 10-14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "w1 = 1000.\t#ft\n", + "w2 = 2000.\t#ft\n", + "r = 700.\t#gpm\n", + "d = 10. \t#days\n", + "q = 2\t #ft\n", + "u = 1.87*((3.4*10**-5)/(3.2*10**4))*(d**6/d)\t#ft\n", + "W = 7.94\t#ft\n", + "p = 114.6*(7*10**2)*W/(3.2*10**4)\t#ft\n", + "U = 1.87*((3.4*10**-5)/(3.2*10**4))*(4*d**6/d)\t#ft\n", + "Wu = 6.55\t#ft\n", + "P = 114.6*(7*10**2)*Wu/(3.2*10**4)\t#ft\n", + "R = 54\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "W1 = R+p+P\t#ft\n", + "D = R+q*p\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the expected drawndown the first well is pumped at a rate = %.0f ft'%(W1)\n", + "print 'the drawdown in each well all the three are pupped at a rate = %.0f ft'%(D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the expected drawndown the first well is pumped at a rate = 90 ft\n", + "the drawdown in each well all the three are pupped at a rate = 94 ft\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.3 Page No : 10-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "g = 20\t#ft\n", + "k = 10**-1\t#cm/sec\n", + "g1 = 3.28*10**-3\t#fps\n", + "w = 2\t#ft\n", + "w1 = 30\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "Q = (1./2)*(g1)*((g**2)-(2**2))/(w1)\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'the flow into a foot of gallery = %.1e cfs'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the flow into a foot of gallery = 2.2e-02 cfs\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch11.ipynb b/Water_and_Wastewater_Engineering/ch11.ipynb new file mode 100644 index 00000000..66e86589 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch11.ipynb @@ -0,0 +1,227 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d97ef885e741db81adc7f9703c96d41639df30eb154ebfc4a587f4f7ba56c3d8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11 : Surface-Water Collection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1 Page No : 11-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "s = 20.\t#mph\n", + "t = 90.\t#min\n", + "w = 1.31\t#ft\n", + "h = 7.5\t#miles\n", + "h1 = 0.22\t#ft\n", + "t1 = 1100.\t#min\n", + "t2 = 6.0\t#min\n", + "p = 32.2\t#ft\n", + "l = 5.12\t#length\n", + "l1 = 2.8\t#length\n", + "p1 = 1400.\t#ft\n", + "d = 73.\t#depth\n", + "h3 = 2.06\t#ft\n", + "e = 173.0\t#ft\n", + "hi = 0.2\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "W = s*w\t#mph\n", + "hs = h1*((W)**2/p)**0.53*h**0.47\t#ft\n", + "Ts = t2*(W/p)**0.44*(h/p)**0.28\t#sec\n", + "Td = t1*h/(p*Ts)\t#min\n", + "Ls = l1/(l*(Ts)**2)\t#ft\n", + "D = d/(l*(Ts)**2)\t#ft\n", + "H = (W)**2*(h*(1/(p1*d)))\t#ft\n", + "hr = h3*l1\t#ft\n", + "M = e+hi+hr\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the overwater wind speed = %.0f mph'%(W)\n", + "print 'the significant wave height = %.1f ft'%(hs)\n", + "print 'the significant wave period = %.1f sec'%(Ts)\n", + "print 'the minimum wind duration required to reach the significant wave height = %.0f min'%(Td)\n", + "print 'the significant wave lenght adn steepness = %.3f ft'%(Ls)\n", + "print 'the reservoir depth ratio = %.1f ft'%(D)\n", + "print 'the wind tide or set up = % f ft'%(H)\n", + "print 'the run up = %.1f ft'%(hr)\n", + "print 'the maximum elevation reached by the waves = %.1f ft'%(M)\n", + "\n", + "# rounding off error. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the overwater wind speed = 26 mph\n", + "the significant wave height = 2.9 ft\n", + "the significant wave period = 3.6 sec\n", + "the minimum wind duration required to reach the significant wave height = 70 min\n", + "the significant wave lenght adn steepness = 0.041 ft\n", + "the reservoir depth ratio = 1.1 ft\n", + "the wind tide or set up = 0.050375 ft\n", + "the run up = 5.8 ft\n", + "the maximum elevation reached by the waves = 179.0 ft\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 Page No : 11-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "g = 264\t#quartz\n", + "p = 0.39\t#percent\n", + "\t\n", + "#CALCULATIONS\n", + "S = (1-p)*(g-1)\t#in\n", + "\t\n", + "#RESULTS\n", + "print 'the hydraulic gradient and seepage velocity = %.2f in'%(S)\n", + "#incorrect answer in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the hydraulic gradient and seepage velocity = 160.43 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 Page No : 11-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "w = 40\t#ft\n", + "k = 2*10**-3\t#cm/sec\n", + "p = 3.28*10**-3\t#cfs\n", + "h = 6.47*10**5\t#gpd\n", + "p1 = 0.433\t#ft\n", + "m = 9\t#ft\n", + "delh = w/(18*9)\t#in\n", + "k1 = 4.94*10**-4\t#cm/sec\n", + "\t\n", + "#CALCULATIONS\n", + "Q = k*p*w*(9./18)\t#cfs\n", + "Q1 = Q*h\t#gpd/ft width\n", + "P = (1-8./18)*w*p1\t#Psig\n", + "H = k1/k\t#in\n", + "\t\n", + "#RESULTS\n", + "print 'the seepage through each foot width of the foundation = %.0f gpd/ft/ width'%(Q1*10)\n", + "print 'the excess hydrostatic pressure on the upstream side of the bottom of the sheet pilling = %.2f Psig'%(P)\n", + "print 'the maximum hydraulic gradient and its relations to the coeeficent = %.2f in'%(H)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the seepage through each foot width of the foundation = 849 gpd/ft/ width\n", + "the excess hydrostatic pressure on the upstream side of the bottom of the sheet pilling = 9.62 Psig\n", + "the maximum hydraulic gradient and its relations to the coeeficent = 0.25 in\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.4 Page No : 11-13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "d = 120\t#ft\n", + "w = 16\t#ft\n", + "d1 = 120/0.8\t#ft\n", + "p = 60*0.8\t#ft\n", + "h = 2\t#ft\n", + "v = 18.74*0.8\t#ft\n", + "s = 95.23\t#ft\n", + "s1 = 0.8\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "W = d-h*p\t#ft\n", + "S = s*s1\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'in succession from the intersection of the upstream slop = %.2f ft'%(S)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "in succession from the intersection of the upstream slop = 76.18 ft\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch12.ipynb b/Water_and_Wastewater_Engineering/ch12.ipynb new file mode 100644 index 00000000..752b1d0d --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch12.ipynb @@ -0,0 +1,236 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:14bd2b02adda106b7feb66c435425687c9047c3f660e50bfc5fca311a000dc14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 : Water Transmission" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 Page No : 12-9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "c = 100\t#in\n", + "a = 10\t#in\n", + "Q = 0.976\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "G = a*Q\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the graphical basic = %.2f ft'%(G)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the graphical basic = 9.76 ft\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2 Page No : 12-9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "a = 27.6\t#sq ft\n", + "h = 1.37\t#ft\n", + "d = 1.53*(27.9)**0.38*(1.36)**0.24\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "R = d/4\t#ft\n", + "A = (math.pi*d**2)/4\t#sq ft\n", + "\t\n", + "#RESULTS\n", + "print 'The diameter hydraulics radius and area of the hydraulically equivalent circular conduit = %.1f sq ft'%(A)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diameter hydraulics radius and area of the hydraulically equivalent circular conduit = 26.7 sq ft\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3 Page No : 12-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "h1 = 13.5\t#ft\n", + "h2 = 19.0\t#ft\n", + "h3 = 27.5\t#ft\n", + "c1 = 2.0*10**4\t#ft\n", + "c2 = 2.1*10**4\t#ft\n", + "c3 = 2.2*10**4\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "H = h1+h2+h3\t#ft\n", + "C = c1+c2+c3\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the most economical distributions of the available head = %.1e ft'%(C)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the most economical distributions of the available head = 6.3e+04 ft\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4 Page No : 12-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "p = 60\t#in\n", + "h = 20\t#percent\n", + "a = 1000\t#ft\n", + "h1 = 40\t#percent\n", + "c = 0.5\t#ft\n", + "p1 = 14.3\t#ft\n", + "p2 = 6.1\t#ft\n", + "d = 11.7*10**-2\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "deltaV = 26.7 - 18.3 # fps\n", + "eq = 3.9*10**-2 * math.sqrt(deltaV/c)*(0.426)**0.356\n", + "D = p*eq\t #ft\n", + "\t\n", + "#RESULTS\n", + "print 'the air valve with a discharge the change in slop = %.2f in.'%(D)\n", + "\n", + "# note : slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the air valve with a discharge the change in slop = 7.08 in.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.5 Page No : 12-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "p = 90.\t #deg\n", + "h = 48.\t #in\n", + "p1 = 100.\t#psig\n", + "P = (1/2*math.pi)*h**2*p1*0.7071\t#lb\n", + "r = 3000./54-31\t#ft\n", + "l = 170.\t#in\n", + "b = 6.5*10**-6\t#ft\n", + "w = 46. \t#ft\n", + "w1 = 1000.\t #ft\n", + "\t\n", + "#CALCULATIONS\n", + "s = b*w*30*10**6\n", + "D = (1./4*math.pi)*h**2*p1\t#lb\n", + "P = (r)*h**2\t#lb\n", + "T = math.pi*h*(1/4)*s\t#lb\n", + "T1 = (1./2)*l\t#tons\n", + "Del = b*w*w1\t#ft per\n", + "\t\n", + "#RESULTS\n", + "print 'The expansion and contraction of the steel line can be as great as = %.1f ft per 1000ft of length, if unrestrained'%(Del)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The expansion and contraction of the steel line can be as great as = 0.3 ft per 1000ft of length, if unrestrained\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch13.ipynb b/Water_and_Wastewater_Engineering/ch13.ipynb new file mode 100644 index 00000000..b7870ef8 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch13.ipynb @@ -0,0 +1,229 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ee62a0db1d14e1ef03e248f6d6042a1fdd2b0d432d5212bd4ea1bd700bb94f91" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13 : Water Distribution" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.2 Page No : 13-14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "p1 = 7.8\t#ft\n", + "p2 = 6.0\t#ft\n", + "p3 = 7.4\t#ft\n", + "p4 = 6.5\t#ft\n", + "p = 7.6\t#ft\n", + "h = 1.0\t#ft\n", + "h1 = 6.7\t#ft\n", + "p5 = 3.3\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "D = p1-p2\t#mgd\n", + "D1 = p1-p3\t#mgd\n", + "D2 = p-p4\t#mgd\n", + "D3 = p4+h\t#mgd\n", + "D4 = h1-p5\t#mgd\n", + "D5 = 2.0 - 1.6\n", + "\t\n", + "#RESULTS\n", + "print \"1 . Deficiency : %.1f mgd\"%D\n", + "print \"2 . Deficiency : %.1f mgd\"%D1\n", + "print \"3 . Deficiency : %.1f mgd\"%D2\n", + "print \" Net added capacity = %.1f mgd\"%(1.2 - 0.2)\n", + "print \" Reinforced capacity : %.1f mgd\"%(D3)\n", + "print \"4 . Deficiency : %.1f mgd\"%D4\n", + "print \" Net added capacity : %.1f mgd\"%(p5 - 0.3)\n", + "print \" Reinforced capacity : %.1f mgd\"%(p5+3)\n", + "print \"5. Deficiency : %.1f mgd\"%D5 \n", + "print \" Net added capacity : %.1f mgd\"%(0.6 - 0.2 )\n", + "print \" Reinforced capacity : %.1f mgd\"%(1.6+0.4)\n", + "print 'the demand is taken = %.1f mgd'%(D3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1 . Deficiency : 1.8 mgd\n", + "2 . Deficiency : 0.4 mgd\n", + "3 . Deficiency : 1.1 mgd\n", + " Net added capacity = 1.0 mgd\n", + " Reinforced capacity : 7.5 mgd\n", + "4 . Deficiency : 3.4 mgd\n", + " Net added capacity : 3.0 mgd\n", + " Reinforced capacity : 6.3 mgd\n", + "5. Deficiency : 0.4 mgd\n", + " Net added capacity : 0.4 mgd\n", + " Reinforced capacity : 2.0 mgd\n", + "the demand is taken = 7.5 mgd\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.3 Page No : 13-15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "w = 500. \t#ft\n", + "p = 20. \t#psig\n", + "h = 40. \t#psig\n", + "h1 = 1000. \t#in\n", + "q = 1250.\t#ft\n", + "g = 2.308/0.75\t#ft\n", + "g1 = 2.308/1.00\t#ft\n", + "s = 5200.\t#gpm\n", + "a = 250. \t#gpm\n", + "\t\n", + "#CALCULATIONS\n", + "H = (h1-(1./2)*(w))\t#ft\n", + "H1 = (h-p)*g\t#percent\n", + "Q = (q-(1./2)*(w))\t#ft\n", + "Q1 = (h-p)*g1\t#percent\n", + "S = s/a\t#gpm\n", + "\t\n", + "#RESULTS\n", + "print 'the number of standard fire streams = %.1f gpm'%(S)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the number of standard fire streams = 20.8 gpm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.6 Page No : 13-21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "h1 = 2.1*3\t#ft\n", + "h2 = 2.1\t#ft\n", + "h = 8.4\t#ft\n", + "p = 1000.\t#ft\n", + "h3 = 5.7\t#ft\n", + "h4 = 4.2*3\t#ft\n", + "q = 4.2\t#ft\n", + "s = 1.68\t#ft\n", + "q1 = 1.33\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "A = p*h/h2\t#ft\n", + "B = p*(h3+h4)/q\t#ft\n", + "C = p*(h1+h2)/s\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the equilent pipe for the Hazen willians coefficent = %.0f ft'%(A)\n", + "print 'the equilent pipe for the Hazen willians coefficent = %.0f ft'%(round(B,-1))\n", + "print 'the equilent pipe for the Hazen willians coefficent = %.0f ft'%(C)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the equilent pipe for the Hazen willians coefficent = 4000 ft\n", + "the equilent pipe for the Hazen willians coefficent = 4360 ft\n", + "the equilent pipe for the Hazen willians coefficent = 5000 ft\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.8 Page No : 13-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "d = 10\t#hr\n", + "p = 50000\t#in\n", + "a = 7.5\t#mgd\n", + "w = 0.75\t#mg\n", + "s = 5.03\t#mg\n", + "\t\n", + "#CALCULATIONS\n", + "S = s/w\t#mg\n", + "P = S-s\t#mg\n", + "\t\n", + "#RESULTS\n", + "print 'a steady gravity supply equal to maximum daily = %.2f mg'%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a steady gravity supply equal to maximum daily = 1.68 mg\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch14.ipynb b/Water_and_Wastewater_Engineering/ch14.ipynb new file mode 100644 index 00000000..2498efee --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch14.ipynb @@ -0,0 +1,550 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b3817b87ab26f1af6cc1ad821b24c26f6f6f9d19671fc5e655b736548ecb040a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14 : Wastewater Flows" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1 Page No : 14-4" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "n = 0.013\t#ft\n", + "s = 4.90\t#ft\n", + "v = 0.590\t#ft\n", + "d = 0.463\t#ft\n", + "w = 3.9*10**-2\t#ft\n", + "p = 1.696\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "V = s*v\t#fps\n", + "Q = s*d\t#cfs\n", + "N = (w*p)**2*1000\t#percent\n", + "\t\n", + "#RESULTS\n", + "print 'the velocity of flow and rate of discharge = %.2f percent'%(N)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the velocity of flow and rate of discharge = 4.38 percent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2 Page No : 14-7" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "v = 1.34\t#fps\n", + "s = 3.7*10**-3\t#fps\n", + "k = 0.8\t#ft\n", + "r = 20\t#ft\n", + "k1 = 0.04\t#ft\n", + "v = 3.0\t#fps\n", + "v1 = 5.0\t#fps\n", + "d = 10**-1\t#ft\n", + "d1 = 1.34\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "V = round((1.49/1.3*10**-2)*(1./4)**(1./6)*(k1*d/30.48*(2.65-1)/1)**(1./2) * 10000,2)\n", + "# for v = 3\n", + "v1 = 3.\n", + "D1 = d*(v1/V)**2\n", + "\n", + "# for v = 5\n", + "v2 = 5\n", + "D2 = d*(v2/V)**2\n", + "\t\n", + "#RESULTS\n", + "print 'For velocity = %d, the gradient at the which coarse quartz = %.1f cm'%(v1,D1) #incorrect answer in the textbook\n", + "print \"For velocity = %d, the gradient at the which coarse quartz = %.1f cm\"%(v2,D2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For velocity = 3, the gradient at the which coarse quartz = 0.5 cm\n", + "For velocity = 5, the gradient at the which coarse quartz = 1.4 cm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3 Page No : 14-10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\t\n", + "#initialisation of variables\n", + "v = 2.5\t#fps\n", + "q = 0.873\t#cfs\n", + "s = 5.20\t#percent\n", + "a = 0.252\t#ft\n", + "r = 0.684\t#ft\n", + "r1 = 1.46\t#ft\n", + "v1 = 0.776\t#ft\n", + "q1 = 0.196\t#ft\n", + "n = 0.78\t#ft\n", + "R = 0.939\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "V = v1*v\t#fps\n", + "Q = q1*q\t#cfs\n", + "R1 = r1*s\t#percent\n", + "Vs = R*v\t#ft\n", + "N = n*Vs\t#fps\n", + "Qs = a*R*q\t#cfs\n", + "N1 = n*Qs\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'the required grades and associated velocity and rates = %.3f cfs'%(V)\n", + "print 'the depth and a grade = %.3f cfs'%(Q)\n", + "print 'the self cleaning flow = %.3f cfs'%(N1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the required grades and associated velocity and rates = 1.940 cfs\n", + "the depth and a grade = 0.171 cfs\n", + "the self cleaning flow = 0.161 cfs\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4 Page No : 14-11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "Q = 0.873\t#cfs\n", + "s = 5.20\t#percent\n", + "d = 0.161\t#cfs\n", + "q1 = 0.185\t#ft\n", + "d2 = 2.5\t#ft\n", + "v = 0.91\t#ft\n", + "s1 = 1.70\t#ft\n", + "s3 = 1.46\t#ft\n", + "w = 0.185\t#ft\n", + "d1 = 0.30\t#ft\n", + "v1 = 0.732\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "q = d/Q\t#cfs\n", + "Vs = v*d2\t#fps\n", + "Ss = s1*s\t#percent\n", + "Va = v1*d2\t#fps\n", + "Ss1 = s3*s\t#percent\n", + "\t\n", + "#RESULTS\n", + "print 'the depth and velocity of flow and the required slop = %.1f percent'%(Ss1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the depth and velocity of flow and the required slop = 7.6 percent\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5 Page No : 14-13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "d1 = 0.67\t#ft\n", + "h1 = 2.00\t#ft\n", + "h2 = 4.04\t#ft\n", + "hv1 = 0.062\t#ft\n", + "hv2 = 0.254\t#ft\n", + "d = 0.19\t#ft\n", + "h = 0.2\t#ft\n", + "h1 = 0.04\t#ft\n", + "q = 0.644\t#ft\n", + "q1 = 0.65\t#ft\n", + "v = 0.92\t#ft\n", + "d2 = 6.5\t#ft\n", + "v1 = 3.69\t#ft\n", + "d3 = 0.542\t#ft\n", + "hv3 = 0.21\t#ft\n", + "delv = 0.15\t#ft\n", + "d4 = 0.02\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "H = d1+hv1\t#ft\n", + "H1 = d1+hv2\t#ft\n", + "he = h*d\t#ft\n", + "hi = d+h1\t#ft\n", + "H2 = d3+hv3\t#ft\n", + "he1 = h*delv\t#ft\n", + "S = d4+h1\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the required slope = %.2f ft'%(hi)\n", + "print 'the lower sewer and the invert drop in the transition = %.2f ft'%(S)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the required slope = 0.23 ft\n", + "the lower sewer and the invert drop in the transition = 0.06 ft\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6 Page No : 14-17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "q = 60\t#cfs\n", + "D = 4\t#ft\n", + "w = 0.177\t#ft\n", + "s = 0.59\t#ft\n", + "h = 4.0\t#ft\n", + "d1 = 1.0\t#ft\n", + "v = 0.90\t#ft\n", + "d1 = 0.42\t#ft\n", + "h1 = 6.0\t#ft\n", + "h2 = 1.5\t#ft\n", + "dl = 1.3\t#ft\n", + "p = 0.41\t#ft\n", + "u = 0.8\t#ft\n", + "u1 = 3.2\t#ft\n", + "y = 0.45\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "H = s*D\t#ft\n", + "d2 = d1*D\t#ft\n", + "V = v*D\t#ft\n", + "P = p*D\t#ft\n", + "D1 = y*D\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the critical depth = %.1f ft'%(H)\n", + "print 'the alternate stages for an energy = %.1f ft'%(V)\n", + "print 'the alternate stages for an energy head = %.1f ft'%(P)\n", + "print 'the lower alternate stage with upper alternate stage = %.1f ft'%(D1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the critical depth = 2.4 ft\n", + "the alternate stages for an energy = 3.6 ft\n", + "the alternate stages for an energy head = 1.6 ft\n", + "the lower alternate stage with upper alternate stage = 1.8 ft\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7 Page No : 14-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "d = 106\t #cfs\n", + "q = 400.\t#cfs\n", + "d1 = 0.40\t#cfs\n", + "w = 10\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "D = d/q\t#cfs\n", + "D1 = d1*w\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'the water level in this well rises = %.f ft'%(D1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the water level in this well rises = 4 ft\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.8 Page No : 14-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\t\n", + "#initialisation of variables\n", + "Q = (400/78.5*math.sqrt(10*32.2))**2\t#ft\n", + "N = 0.012\t#ft\n", + "d = 0.47\t#ft\n", + "q = 10\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "D = d*q\t#ft\n", + "\n", + "#RESULTS\n", + "print 'the water surface in the sewer when it is flowing at maximum capacity = %.1f ft'%(D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the water surface in the sewer when it is flowing at maximum capacity = 4.7 ft\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9 Page No : 14-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variable\n", + "g = math.sqrt(3)\t#ft\n", + "d = 5.67\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "C = g*d\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'The rate of propagation of a discontinuous surge = %.1f ft'%(C)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate of propagation of a discontinuous surge = 9.8 ft\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.10 Page No : 14-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#initialisation of variables\n", + "Q1 = 30\t#cfs\n", + "Q2 = 16\t#cfs\n", + "a = 32\t#sq ft\n", + "r = 1.6\t#ft\n", + "i = 10**-4\t#ft\n", + "n = 1.25*10**-2\t#ft\n", + "h2 = 0.50\t#ft\n", + "c = 3.33\t#ft\n", + "h1 = 5.20\t#ft\n", + "l = 72\t#ft\n", + "s = 12320\t#ft\n", + "\n", + "#CALCULATIONS\n", + "L = s-l\t#ft\n", + "h1 = 0.49022 - 8.04*10**-5*l\t\n", + "#RESULTS\n", + "print 'the forchheimer s methos = %.0f ft'%(L)\n", + "print (Q1**2-Q2**2)/(2*32.2*a**2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the forchheimer s methos = 12248 ft\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11 Page No : 14-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "q = 1.0\t#cfs\n", + "g = 2.0\t#percent\n", + "g1 = 5.6\t#percent\n", + "r = 0.015\t#cfs\n", + "w = 90.\t#percent\n", + "Q = 10*0.9*q\t#ft\n", + "p = 0.10\t#ft\n", + "h = 3.48*g1**(1./3)\t#ft\n", + "i = 5.6*10**-2\t\n", + "Q0 = 1.\n", + "s = 2*10**-2\n", + "n = 1.5*10**-2\n", + "\n", + "#CALCULATIONS\n", + "q_l = round(1.87*i**0.569*(Q0/(math.sqrt(s)/n))**0.563,2)\n", + "l = q_l * 100\n", + "Q = 10*0.9*Q0\n", + "d = (3.48*32.2**(1./3)) * (q_l)**(2./3)\n", + "\n", + "#RESULTS\n", + "print 'The maximum depth of flow in the gutter = %.1f in'%(d)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum depth of flow in the gutter = 2.4 in\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch15.ipynb b/Water_and_Wastewater_Engineering/ch15.ipynb new file mode 100644 index 00000000..5101dc59 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch15.ipynb @@ -0,0 +1,221 @@ +{ + "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": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch16.ipynb b/Water_and_Wastewater_Engineering/ch16.ipynb new file mode 100644 index 00000000..2b395898 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch16.ipynb @@ -0,0 +1,64 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3c3856bc217e5f495e078375001a6745906e09c45dfca2c5d1fe105500ef94b8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 16 : Machinery and Equipment" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.2 Page No : 16-11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p = 500\t#ft\n", + "p1 = 6\t#in\n", + "t = 500\t#cfm\n", + "p2 = 7\t#psig\n", + "P = p2+14.7\t#psia\n", + "T = 520*(P/14.7)**0.283\t#F\n", + "f = 0.048*p1**0.027/(t)**0.148\t#in\n", + "\t\n", + "#CALCULATIONS\n", + "delP = 20.*10**-3*p*T*(t)**2/(38*10**3*P*p1**5)\t#psia\n", + "\t\n", + "#RESULTS\n", + "print 'the pressure drop = %.2f psia'%(delP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the pressure drop = 0.23 psia\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch2.ipynb b/Water_and_Wastewater_Engineering/ch2.ipynb new file mode 100644 index 00000000..d099aba5 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch2.ipynb @@ -0,0 +1,374 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:af736a02c1f2601e59ec790e080728198307e957dded2cfbca28aa79de429565" + }, + "nbformat": 3, + "nbformat_minor": 0, + "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": {} + } + ] +}
\ No newline at end of file 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": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch4.ipynb b/Water_and_Wastewater_Engineering/ch4.ipynb new file mode 100644 index 00000000..c47b5c94 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch4.ipynb @@ -0,0 +1,287 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:69dc81267d132717cceb5ed01afab9cdbb37bfc431d05698243961d6dd4f91a3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Information Analysis" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 4-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import *\n", + "\n", + "#initialisation of variables\n", + "y_bar = 19.5\t#in\n", + "x = 396.8\t#in\n", + "n = 6\t#in\n", + "y1 = 2.20\t#in\n", + "x1 = 51.14\t#in\n", + "p = 5.64\t#in\n", + "ob_y = array([44,20,24,14,12,3])\n", + "ob_x = array([5.3,3.5,3.,1.2,0.48,-0.26])\n", + "\n", + "#CALCULATIONS\n", + "Beta = round((x-n*(y_bar)*(y1))/(x1-n*(y1)**2),2)\n", + "y = around(y_bar + Beta*(ob_x - 2.2),decimals=1)\n", + "Ri = ob_y - y\n", + "sumRi = sum(Ri)\n", + "\n", + "#RESULTS\n", + "print \"Beta B = %.2f\"%Beta\n", + "print \"residuals Ri are : \" ,Ri\n", + "print 'the method of leate squares = %.2f minimum'%(sumRi)\n", + "\n", + "# note : answer is slighty different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Beta B = 6.31\n", + "residuals Ri are : [ 4.9 -7.7 -0.5 0.8 3.4 -1. ]\n", + "the method of leate squares = -0.10 minimum\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 Page No : 4-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import *\n", + "\n", + "#initialisation of variables\n", + "a = 12.\t #in\n", + "h = 121.\t#in\n", + "p = 11. #in\n", + "s = 220.\t#in\n", + "observed_time = array([11,12,13,14,15,16,17,18,19,20,21])\n", + "i = array([1,2,3,4,5,6,7,8,9,10,11])\n", + "observed_magnitude = array([2,4,6,8,10,12,14,16,18,20,22])\n", + "y_uy = array([-10,-8,-6,-4,-2,0,2,4,6,8,10])\n", + "\n", + "#CALCULATIONS\n", + "B = a/p*(h-1)*s\t#per unit\n", + "n = len(i)\n", + "i_6 = i - (n+1)/2\n", + "i_6_y_uy = (i - (n+1)/2)*y_uy\n", + "\n", + "#RESULTS\n", + "print \"Observed Time \", observed_time\n", + "print \"Order,i \", i\n", + "print \"[i-(n+1)/2] \", i_6\n", + "print \"Observed magnitude \", observed_magnitude\n", + "print \"Deviation from mean \", y_uy\n", + "print \"[i-(n+1)/2](y-uy) \", i_6_y_uy \n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Observed Time [11 12 13 14 15 16 17 18 19 20 21]\n", + "Order,i [ 1 2 3 4 5 6 7 8 9 10 11]\n", + "[i-(n+1)/2] [-5 -4 -3 -2 -1 0 1 2 3 4 5]\n", + "Observed magnitude [ 2 4 6 8 10 12 14 16 18 20 22]\n", + "Deviation from mean [-10 -8 -6 -4 -2 0 2 4 6 8 10]\n", + "[i-(n+1)/2](y-uy) [50 32 18 8 2 0 2 8 18 32 50]\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No : 4-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "a = 4404.\t#ft\n", + "q = 9.\t#ft\n", + "mu = 12.\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "F = math.sqrt(a/q)\t#ft\n", + "CF = F/mu*100\t#percent\n", + "\t\n", + "#RESULTS\n", + "print 'the coefficient of fluctuation is = %.0f percent'%(CF)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the coefficient of fluctuation is = 184 percent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 Page No : 4-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "h2 = 5\t#in\n", + "x = 3.72\t#in\n", + "x1 = 1.28\t#in\n", + "\t\n", + "#CALCULATIONS\n", + "H = h2*x1/x\t#in\n", + "\t\n", + "#RESULTS\n", + "print 'the either side of the center of the scale = %.2f in'%(H)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the either side of the center of the scale = 1.72 in\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 Page No : 4-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p = 80\t#in\n", + "q = 20\t#in\n", + "\t\n", + "#CALCULATIONS\n", + "K = p+q\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the moments of the arithmetically normal frequency curve = %.0f ft'%(K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the moments of the arithmetically normal frequency curve = 100 ft\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No : 4-38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "g = 3.2541\t#in\n", + "g1 = 3.46\t#in\n", + "m = 0.5390\t#ft\n", + "h = 2./99\t#ft\n", + "p = 1.52\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "L = math.sqrt(g*h)\t#in\n", + "mu = g1*p\t#in\n", + "M = g1/p\t#percent\n", + "\t\n", + "#RESULTS\n", + "print 'the points necessary to plot the straigt line of fit on math.log probability = %.2f percent'%(M)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the points necessary to plot the straigt line of fit on math.log probability = 2.28 percent\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch5.ipynb b/Water_and_Wastewater_Engineering/ch5.ipynb new file mode 100644 index 00000000..1946a849 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch5.ipynb @@ -0,0 +1,311 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d903acbbd579075c20da58df0664f6b615313ecea1f356fbfe69d66ce9759ff7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Water and Wastewater Volumes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page No : 5-6" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "t1 = 5.25\t#yr\n", + "t2 = 10.00\t#yr\n", + "yi = 171000\t#in\n", + "ye = 111000\t#in\n", + "yt = 5.23300\t#in\n", + "yl = 5.04532\t#in\n", + "yn = 31500\t#in\n", + "ym = 0.09853\t#in\n", + "tm = 9.25\t#yr\n", + "tn = 10.00\t#yr\n", + "\t\n", + "#CALCULATIONS\n", + "T = t1/t2\t#yr\n", + "T1 = tm/tn\t#yr\n", + "Y = yi-ye\t#in\n", + "Yt = yt-yl\t#in\n", + "\t\n", + "#RESULTS\n", + "print 'the fifth intercensal year = %.3f yr'%(T)\n", + "print 'the ninth postcensal year = %.3f yr'%(T1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the fifth intercensal year = 0.525 yr\n", + "the ninth postcensal year = 0.925 yr\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No : 5-8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import Symbol\n", + "import math\n", + "\n", + "#initialisation of variables\n", + "y0 = 30000\t#in\n", + "y1 = 172000\t#in\n", + "y2 = 292000\t#in\n", + "a = 172\t #ft\n", + "p = 30.\t #ft\n", + "y = 292\t #ft\n", + "q = 322000\t#ft\n", + "g = 313\t #ft\n", + "n = 0.05\t#ft\n", + "d = -2.442\t#ft\n", + "t = Symbol('t')\n", + "\n", + "#CALCULATIONS\n", + "L = (2*p*a*y2-(a)**2*q)/(p*y-(a)**2)\t#moreover\n", + "m = (g-p)/p\t#ft\n", + "N = n*d\t#in\n", + "Y = g/(1+9.43*math.exp(N)*t)\t#in\n", + "\t\n", + "#RESULTS\n", + "print 'the saturation populations = %.2f moreover'%(round(L,-3))\n", + "print 'the coefficients = %.3f in'%(N)\n", + "print 'the equation of a logistic curve = ',(Y)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the saturation populations = 313000.00 moreover\n", + "the coefficients = -0.122 in\n", + "the equation of a logistic curve = 313/(8.34611446180163*t + 1)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 Page No : 5-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "p = 100000\t#in\n", + "d = 150.\t#in\n", + "h = 1000000.\t#in\n", + "a1 = 2.0\t#draft\n", + "a2 = 3.0\t#draft\n", + "a3 = 1.6\t#draft\n", + "m = 1.5\t #in\n", + "q = 2.5\t #in\n", + "v = 1020.\t#in\n", + "w = 100. \t#in\n", + "t = 0.01\t#in\n", + "v1 = 13.2\t#mgd\n", + "\t\n", + "#CALCULATIONS\n", + "A = d*p/h\t#mgd\n", + "M = m*A\t#mgd\n", + "M1 = q*A\t#mgd\n", + "V = v*math.sqrt(w)*(1-t*math.sqrt(w))\t#gpm\n", + "D = M+v1\t#mgd\n", + "L = a1*A\t#mgd\n", + "L1 = (4./3)*M\t#max\n", + "H = a2*A\t#mgd\n", + "H1 = (4./3)*M1\t#max\n", + "F = a3*A\t#mgd\n", + "F1 = (1.6)*15\t#max\n", + "\t\n", + "#RESULTS\n", + "print 'the resulting capacities of the four system = %.f max'%(F1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the resulting capacities of the four system = 24 max\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 Page No : 5-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "r = 48\t#in\n", + "A = 450\t#gpd/acre\n", + "B = 8000.\t#gpd/mile\n", + "S = round(5280./350)\t#manholes/mile\n", + "\n", + "#CALCULATIONS\n", + "C = (B-S*100)/12\t#gpd/mile\n", + "\n", + "#RESULTS\n", + "print 'the ground a quarter of it eventually = %.2f gpd/mile'%(C)\n", + "\n", + "# note : answer in book is wrong. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the ground a quarter of it eventually = 541.67 gpd/mile\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 Page No : 5-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p1 = 20.\t#ft\n", + "p2 = 30.\t#ft\n", + "w = 5. \t#person\n", + "s = 17800.\t#in\n", + "h = 1200.\t#in\n", + "q = 100. \t#in\n", + "i = 1.\t#in\n", + "\t\n", + "#CALCULATIONS\n", + "S = p1*p2*i*s/(h*w)\t#gpcd\n", + "P = (q*w*10/S)\t#percent\n", + "\n", + "\n", + "#RESULTS\n", + "print 'the degree of separation of stormwater = %.0f percent'%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the degree of separation of stormwater = 3 percent\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8 Page No : 5-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "s = 105\t#gpcd\n", + "m = 315\t#gpcd\n", + "m1 = 35\t#gpcd\n", + "Q1 = 360\t#gpcd\n", + "Q2 = 30\t#gpcd\n", + "p1 = 20\t#pecent\n", + "p2 = 15\t#persons/acer\n", + "D = 21\t#persons/acre\n", + "I = 2000\t#gpd/acre\n", + "\t\n", + "#CALCULATIONS\n", + "A = D*(s+Q2)+I\t#gpd/acre\n", + "R = D*(m+Q2)+I\t#gpd/acre\n", + "L = D*(m1+Q2)+I\t#gpd/acre\n", + "\t\n", + "#RESULTS\n", + "print 'the average peak and low rates of flow = %.0f gpd/acre'%(round(L,-1))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the average peak and low rates of flow = 3370 gpd/acre\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch6.ipynb b/Water_and_Wastewater_Engineering/ch6.ipynb new file mode 100644 index 00000000..1c733ac1 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch6.ipynb @@ -0,0 +1,163 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:098e6f8caaa45e66f3c3c4de6e35624c980f673866a521599c34dc5ef9f6350c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 : Elements of Hydrology" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1 Page No : 6-6" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "H = 1360\t#ft\n", + "t = 60\t#f\n", + "a = (10**3)*5.5*(10**-3)\t#f\n", + "q = (1.36*10**3)*5.5*(10**-3)\t#f\n", + "s = (4-1.36)*(10**3)*(3.2*10**-3)\t#f\n", + "\t\n", + "#CALCULATIONS\n", + "T = t-q-s\t#F\n", + "T1 = T+3*a\t#F\n", + "\t\n", + "#RESULTS\n", + "print 'the temperature at the mountain top = %.0f F'%(T)\n", + "print 'the temperature on the plain beyond the mountain = %.1f F'%(T1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the temperature at the mountain top = 44 F\n", + "the temperature on the plain beyond the mountain = 60.6 F\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2 Page No : 6-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "t = 60\t#f\n", + "v = 0.52\t#in\n", + "t1 = 80\t#F\n", + "p = 40\t#percent\n", + "v1 = 1.03*0.40\t#in\n", + "w = 8\t#mph\n", + "pa = 29.0\t#in\n", + "p1 = 0.497\t#ft\n", + "q = 1.32*10**-2\t#ft\n", + "r = 0.268\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "E = p1*(1-q*pa)*(1+r*w)*(v-.41)\t#in\n", + "\t\n", + "#RESULTS\n", + "print 'the evaporation for the a day during = %.2e in'%(E)\n", + "\n", + "# note : answer is slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the evaporation for the a day during = 1.06e-01 in\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page No : 6-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "t = 47\t#f\n", + "q = 8000\t#ft\n", + "a = 100\t#ft\n", + "d = 0.10\t#in\n", + "d1 = 7\t#degree days\n", + "s1 = 14000\t#ft\n", + "s2 = 7000\t#ft\n", + "s = 1000\t#ft\n", + "g = 32\t#ft\n", + "h = 17.37\t#ft\n", + "h1 = 1.547\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "T = q+s*(t-g)/3\t#ft\n", + "T1 = t-3*1\t#F\n", + "T2 = (T1+g)/2\t#F\n", + "T3 = d1*d*a\t#sq mile in\n", + "M = h*T3\t#mgd\n", + "M1 = M*h1\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'the upper boundary of the melting zone and temperature at the snow line = %.0f F'%(T1)\n", + "print 'The average temperature of = %d cfs'%(M1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the upper boundary of the melting zone and temperature at the snow line = 44 F\n", + "The average temperature of = 1880 cfs\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch7.ipynb b/Water_and_Wastewater_Engineering/ch7.ipynb new file mode 100644 index 00000000..60a0e5fa --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch7.ipynb @@ -0,0 +1,156 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:42c32bb83042eba762c882953210b5f735a47b1ef07c7eade0266ced5fe417ef" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Rainfall and Runoff" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5 Page No : 7-15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "n = 20. \t#ft\n", + "s = math.sqrt(12676/19.)\t#ft\n", + "c = 45.5\t#ft\n", + "q = 551400\t#ft\n", + "q1 = 12700\t#ft\n", + "h = 8.5\t#ft\n", + "w = s/c\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "D = q/(2*s*q1)\t#cfs\n", + "D1 = D*(1+h/n)\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'the record runoff of a stream draining = %.2f cfs'%(D1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the record runoff of a stream draining = 1.20 cfs\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6 Page No : 7-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "i = 16./(62)**0.66\t#in hr\n", + "q = (16*10**0.31)/(62)**0.66\t#in hr\n", + "c = 1.0\t#max\n", + "C1 = c*(0.01)**0.31\t#in\n", + "C2 = c*(0.1)**0.31\t#in\n", + "x1 = 640\t#cfs\n", + "\t\n", + "#CALCULATIONS\n", + "Y1 = C1*i*c*x1\t#cfs\n", + "Y2 = C2*q*c*x1\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'the time of concentration = %.f cfs'%(Y2)\n", + "\n", + "# rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the time of concentration = 672 cfs\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.8 Page No : 7-25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "d = 163*48.5\t#cfs\n", + "a = 48.5\t#ft\n", + "q = 100\t#cfs\n", + "Q = 45.5*a\t#cfs\n", + "c = 0.57\t#cfs\n", + "v = 1.8\t#cfs\n", + "p = 0.45\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "P = d/(q*math.sqrt(a))\t#percent\n", + "C = Q/(a**0.8*(1+2*a**-0.3))\t#cfs\n", + "d1 = 2.6\t#cfs\n", + "T = (1-p*c+v*c*2)\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'the meyers rating = %.1f percent'%(P)\n", + "print 'the magnitude of the maximum peak flood = %.1f cfs'%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the meyers rating = 11.4 percent\n", + "the magnitude of the maximum peak flood = 2.8 cfs\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch8.ipynb b/Water_and_Wastewater_Engineering/ch8.ipynb new file mode 100644 index 00000000..9aa81db6 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch8.ipynb @@ -0,0 +1,202 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fba260293419dba131eeeb5811674596ce6afa78c83d914173c371599b6aa433" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Storage and Runoff Control" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2 Page No : 8-7" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "a = 0.75\t#ft\n", + "p = 123\t#mg\n", + "v = 100\t#ft\n", + "s = 33\t#mg\n", + "s1 = 67\t#mg\n", + "d = 26.6\t#mgd\n", + "d1 = 0.0477\t#mgd\n", + "q = 0.750\t#gpd/sq mile\n", + "d2 = 365\t#days\n", + "\t\n", + "#CALCULATIONS\n", + "S = p/a\t#mg per sq mile\n", + "Cv = v*s/s1\t#percent\n", + "M = d*d1\t#mgd per sq mile\n", + "D = v*q/M\t#MAF\n", + "D1 = (v*p)/(M*d2)\t#MAF\n", + "R = p/q\t#days\n", + "\t\n", + "#RESULTS\n", + "print 'the use monthly averages rather then daily stream flow = %.0f days'%(R)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the use monthly averages rather then daily stream flow = 164 days\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3 Page No : 8-9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "d = 750000\t#gpd per sq mile\n", + "v = 0.22\t#ft\n", + "a = 1.27\t#ft\n", + "q = 0.30\t#ft\n", + "d1 = 365\t#days\n", + "p = 0.25\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "Q = q*a*d1\t#mg/sq mile\n", + "H = p*a*d1\t#mg/sq mile\n", + "\t\n", + "#RESULTS\n", + "print 'the results obtained by normal analytical procedures and by Hazen s = %.0f mg/sq mile'%(H)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the results obtained by normal analytical procedures and by Hazen s = 116 mg/sq mile\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4 Page No : 8-10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "d = 30.0\t#mgd\n", + "a = 40.0\t#sq miles\n", + "a1 = 1500\t#acres\n", + "r1 = 47.0\t#in\n", + "r2 = 27.0\t#in\n", + "q = 0.9\t#in\n", + "k = 640\t#in\n", + "h = 0.052\t#gpd/sq mile\n", + "\t\n", + "#CALCULATIONS\n", + "Q = r2-(r2+a-r1)*q*a1/(k*a)\t#in\n", + "D = d+a*h\t#mgd\n", + "A = a-(q*a1/k)*(1-(r1-a)/(r2))\t#sq miles\n", + "R = r2+a-r1\t#in\n", + "S = R*q\t#in\n", + "\t\n", + "#RESULTS\n", + "print 'the revised mean annual runoff = %.1f in'%(Q)\n", + "print 'the equivalent mean draft = %.1f mgd'%(D)\n", + "print 'the equivalent land area = %.1f sq miles'%(A)\n", + "print 'the adjusted flowline = %.0f in'%(S)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the revised mean annual runoff = 25.9 in\n", + "the equivalent mean draft = 32.1 mgd\n", + "the equivalent land area = 38.4 sq miles\n", + "the adjusted flowline = 18 in\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6 Page No : 8-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p = 100. \t#ft\n", + "q = 27000.\t #acre-ft\n", + "p1 = 10. \t#ft\n", + "s = 8250.\t#acre-ft\n", + "\t\n", + "#CALCULATIONS\n", + "R = p*s/q\t#percent\n", + "\t\n", + "#RESULTS\n", + "print 'the ratio of peak inflow from fuller values = %d percent'%(R)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the ratio of peak inflow from fuller values = 30 percent\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Water_and_Wastewater_Engineering/ch9.ipynb b/Water_and_Wastewater_Engineering/ch9.ipynb new file mode 100644 index 00000000..17df7377 --- /dev/null +++ b/Water_and_Wastewater_Engineering/ch9.ipynb @@ -0,0 +1,297 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:cb8c1d89336b90a52fb3fc2ac2cf4a335c076b1a2d2bc8b2521052d89b269ec7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 : Groundwater Flow" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1 Page No : 9-11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "t = 10\t#C\n", + "s = 74.2\t#days\n", + "c = 0.01\t#mm\n", + "d = 245\t#mm\n", + "\t\n", + "#CALCULATIONS\n", + "h = s/(d*c)\t#cm\n", + "\t\n", + "#RESULTS\n", + "print 'the high will water at a temperature = %.1f cm'%(h)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the high will water at a temperature = 30.3 cm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2 Page No : 9-13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p1 = 1000\t#ft\n", + "p2 = 50\t#ft\n", + "g = 20\t#ft/mile\n", + "v = 5280.\t#ft\n", + "q = 7.5*10**-6\t#ft\n", + "t = 60\t#F\n", + "k = 2835\t#ft/days\n", + "p = 7.5\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "S = g/v\t#ft\n", + "W = k*(g/v)\t#ft/day\n", + "Q = W*p1*p2*q\t#mgd\n", + "P = k*p\t#ft\n", + "P1 = P*p2\t#mgd\n", + "\t\n", + "#RESULTS\n", + "print 'the velocity of flow = %.2f mgd'%(Q)\n", + "print 'the standard coefficient pf permeability = %.2e mgd'%(P1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the velocity of flow = 4.03 mgd\n", + "the standard coefficient pf permeability = 1.06e+06 mgd\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3 Page No : 9-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\n", + "#initialisation of variables\n", + "p = 40. \t#ft\n", + "d = 56. \t#ft\n", + "d1 = 140. \t#ft\n", + "p1 = 30. \t#ft\n", + "w = 3.28*10**-4\t#fps\n", + "\t\n", + "#CALCULATIONS\n", + "Q = w*(p/d1)*2*d*p\t#cfs\n", + "q = Q/p\t#cfs\n", + "K = w*(p/d1)\t#fps\n", + "x0 = q/(2*math.pi*K)\t#ft\n", + "Z = 2*math.pi*x0\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the yield of the well if the coefficient of permeability = %.1f ft'%(x0)\n", + "print 'the distance of the point of stagnation = %.0f ft'%(Z)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the yield of the well if the coefficient of permeability = 17.8 ft\n", + "the distance of the point of stagnation = 112 ft\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4 Page No : 9-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "p = 5.*10**6\t #ft\n", + "Q = 350. \t#gpm\n", + "x = 225. \t#ft\n", + "u = 10.**-2\t#ft\n", + "g = 1.87\t#ft\n", + "p2 = 7.*10**2\t#ft\n", + "p3 = 10.9\t#ft\n", + "w = 4.0\t#ft\n", + "t = 114.6\t #ft\n", + "d = 10. \t#ft\n", + "p1 = 5. \t#ft\n", + "w1 = 3.2*10**4\t#ft\n", + "W = 21.75\t#ft\n", + "\t\n", + "#CALCULATIONS\n", + "T = t*Q*4/p1\t#gpd/ft\n", + "S = u*(w1)/(g*(p))\t#ft\n", + "U = g*((S)/(T))*(x**2/d)\t#ft\n", + "P = t*(p2)*p3/(T)\t#ft\n", + "U1 = g*((S)/(T))*(1./d)\t#ft\n", + "P1 = t*(p2)*W/(T)\t#ft\n", + "\t\n", + "#RESULTS\n", + "print 'the type curve as if a transparency of the observed data had moved into place over the type = %.0f ft'%(P1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the type curve as if a transparency of the observed data had moved into place over the type = 54 ft\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5 Page No : 9-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\t\n", + "#initialisation of variables\n", + "Q = 350. \t#gpm\n", + "x = 225. \t#ft\n", + "t = 1. \t#min\n", + "p = 1.6\t #ft\n", + "t2 = 10. \t#min\n", + "p2 = 4.5\t#ft\n", + "p3 = 700. \t#gpm\n", + "T = 3.2*10**4\t#gpd/ft\n", + "t0 = 0.3 \t#min\n", + "u = 1.15*10**-5\n", + "\t\n", + "#CALCULATIONS\n", + "S = t0*(T)*t0/((x)**2*1440)\t#ft\n", + "P = ((114.6*p3)/(T))*(-0.5772*2.3*math.log(u))\t#ft\n", + "\n", + "#RESULTS\n", + "print 'A straight line being drawn through the ppints for the higher = %.0f ft'%(P)\n", + "\n", + "# note : book answer is wrong. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "A straight line being drawn through the ppints for the higher = 38 ft\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6 Page No : 9-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\n", + "#initialisation of variables\n", + "h = 4.8\t #ft\n", + "m = 13.4\t #ft\n", + "k = 10.**-1\t #cm/sec\n", + "k1 = 3.28*10**-3\t#fps\n", + "n = 7. \t#ft\n", + "n1 = 11. \t#ft\n", + "q = 1.0*10**-2\n", + "\t\n", + "#CALCULATIONS\n", + "Q = k1*h*n/n1\t#cfs/ft\n", + "Q1 = 2*q*10**2\t#cfs\n", + "\t\n", + "#RESULTS\n", + "print 'A satisfactory orthogonal system the flow of into the collector = %.0f cfs'%(Q1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "A satisfactory orthogonal system the flow of into the collector = 2 cfs\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
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