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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": {} + } + ] +}
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