{ "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": {} } ] }