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diff --git a/Problems_In_Hydraulics/ch5.ipynb b/Problems_In_Hydraulics/ch5.ipynb new file mode 100755 index 00000000..696c90f4 --- /dev/null +++ b/Problems_In_Hydraulics/ch5.ipynb @@ -0,0 +1,422 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:99b4244f7dba40fbe6b1f9647ffae37a043c932f7cb3e888c2ad7aac3ff9563e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Flow in Channels" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page No : 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "h= 2.5 \t#ft depth of water\n", + "a= 45. \t#degrees side slope\n", + "x= 5. \t#ft\n", + "Q= 45. \t#cuses\n", + "v= 2.6 \t#ft/sec velocity\n", + "w= 6.92 \t#ft \n", + "C= 120.\n", + "\n", + "#CALCULATIONS\n", + "b= (Q/(v*h))-h\n", + "p= b+2*(h+math.sqrt(2))\n", + "A= h*w\n", + "m= A/p\n", + "i= (v/(C*math.sqrt(m)))**2\n", + "\n", + "#RESULTS\n", + "print 'Width = %.2f ft'%(b) \n", + "print ' Slope = %.6f '%(i) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Width = 4.42 ft\n", + " Slope = 0.000332 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No : 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "a= 60. \t#degrees sides inclined\n", + "i= 1./1600\n", + "Q= 8.*10**6 \t#gal/hr discharge\n", + "M= 110.\n", + "w= 6.24 \t#lb/ft**3\n", + "\n", + "#CALCULATIOS\n", + "d= ((Q*2**(2./3)*math.sqrt(1./i))/(w*3600*math.sqrt(3)*M))**(3./8)\n", + "b=6.93 \t#ft\n", + "\n", + "#RESULTS\n", + "print 'Diameter = %.f ft'%(d) \n", + "print ' breadth = %.2f ft'%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter = 6 ft\n", + " breadth = 6.93 ft\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 Page No : 71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "g= 32.2 \t#ft/swc**2\n", + "Q= 40. \t#cuses rate\n", + "w= 5.5 \t#ft\n", + "h= 9. \t#in depth\n", + "d= 0.75 \t#ft\n", + "V= 3. \t#ft/sec\n", + "\n", + "#CALCULATIONS\n", + "D= ((Q*2)**2/(g*(w*2)**2))**(1./3)\n", + "v= Q*d/w\n", + "D1= math.sqrt((2*v**2*d/g)+h/64)-(d/2)\n", + "dD= D1-d\n", + "El= -dD+((v**2*(1-(V/v)**2))/(2*g))\n", + "Els= Q*El*62.4/550\n", + "\n", + "#RESULTS\n", + "print 'Critical depth = %.2f ft'%(D)\n", + "print ' Rise in level = %.f ft'%(D1)\n", + "print ' Horse-power lost = %.3f hp'%(Els) \n", + "\n", + "#The answer is a bit different due to rounding off error in textbook\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical depth = 1.18 ft\n", + " Rise in level = 1 ft\n", + " Horse-power lost = 0.961 hp\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 Page No : 77" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "b= 3.5 \t#ft\n", + "H= 2.5 \t#ft\n", + "w= 3. \t#ft depth\n", + "h= 6. \t#ft wide\n", + "g= 32.2 \t#ft/sec**2\n", + "\n", + "#CALCULATIONS\n", + "Q= 3.09*b*H**1.5\n", + "v= Q/(w*h)\n", + "H1= H+(v**2/(2*g))\n", + "Q1= 3.09*b*H1**1.5\n", + "hc= (Q1**2/(b**2*g))**(1./3)\n", + "h2= 0.5*(math.sqrt(hc**2+8*hc**2)-hc)\n", + "dh= h2+b-w\n", + "\n", + "#RESULTS\n", + "print \"Flow rate = %.1f cusecs\"%(Q)\n", + "print \" Flow rate = %d cusecs\"%(Q1)\n", + "print ' maximum depth of water downstream = %.3f ft'%(dh) \n", + "print ' Shooting flow depth at hump = %.3f ft'%(h2) \n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flow rate = 42.8 cusecs\n", + " Flow rate = 45 cusecs\n", + " maximum depth of water downstream = 2.226 ft\n", + " Shooting flow depth at hump = 1.726 ft\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 Page No : 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "m= 60./26\n", + "i= 1./2000\n", + "h1= 3. \t#ft depth\n", + "h2= 5. \t#ft depth\n", + "m1= 10./3\n", + "C= 90. # constant\n", + "l= 500. \t#ft depth\n", + "H= 20. \t#ft broad\n", + "H1= 29.62 \t#ft\n", + "g= 32.2 \t#ft/s**2\n", + "\n", + "#CALCULATIONS\n", + "v= 90*math.sqrt(m*i)\n", + "v1= v*h1/h2\n", + "dh= (i-(v1**2/(C**2*m1)))*l/(1-v1**2/(g*h2))\n", + "h3= h2-dh\n", + "V= h1*v/h3\n", + "\n", + "#RESULTS\n", + "print 'Height of water 1000 ft upstream = %.3f ft'%(h3) \n", + "print ' Height of water upstream = %.3f ft'%(h3) \n", + "\n", + "#The answer is a bit different due to rounding off error in textbook\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Height of water 1000 ft upstream = 4.808 ft\n", + " Height of water upstream = 4.808 ft\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8 Page No : 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "v= 5. \t #ft/sec\n", + "m= 60./26\n", + "i= 1./2000\n", + "h= 5.5 \t#ft\n", + "m1= 110./31\n", + "d= 3. \t #ft\n", + "g= 32.2 \t#ft/sec**2\n", + "\n", + "#CALCULATIONS\n", + "C= v/(math.sqrt(m*i))\n", + "v1= v*d/h\n", + "r= (i-(v1**2/(C**2*m1)))/(1-(v1**2/(g*h)))\n", + "x= 1/r\n", + "\n", + "#RESULTS\n", + "print 'Distance upstream = %.f ft'%(round(x,-1)) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance upstream = 2380 ft\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9 Page No : 81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from numpy import *\n", + "from numpy.linalg import *\n", + "\n", + "#initialisation of variables\n", + "g= 32.2 \t#ft/sec**2\n", + "Q= 12 \t#cuses\n", + "\n", + "#CALCULATIONS\n", + "hc= (Q/(3*math.sqrt(g)))**(2./3)\n", + "vec=roots([1,6,12,8,0,-8.95,-8.95])\n", + "H=vec[2]\n", + "\n", + "#RESULTS\n", + "print 'Critical depth = %.2f ft'%(hc) \n", + "print ' Critical depth = %.2f ft'%(H) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical depth = 0.79 ft\n", + " Critical depth = 0.89 ft\n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "-c:17: ComplexWarning: Casting complex values to real discards the imaginary part\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "Cd= 0.64 # coefficient\n", + "g= 32.2 \t#ft/sec**2\n", + "A= 12.5 \t#ft**2\n", + "H= 24.8 \t#ft\n", + "Q= 3200. \t#cuses\n", + "b= 150. \t#ft wide\n", + "A1= 5.*10**6 # avg surface area\n", + "h= 9. \t#ft\n", + "h1= 6. \t #in\n", + "\n", + "#CALCULATIONS\n", + "N= Q/(Cd*A*math.sqrt(2*g*H))\n", + "H1= (Q/(3.2*b))**(2./3)\n", + "ES= (H1-(h1/12))*A1*h\n", + "\n", + "#RESULTS\n", + "print 'number of siphons = %.f '%(N) \n", + "print ' Extra Storage = %.2e ft**3'%(ES) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "number of siphons = 10 \n", + " Extra Storage = 1.37e+08 ft**3\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
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