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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Chapter 10 : Open Channel Flow\n",
      "    "
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 10.1 Page no 363"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import *\n",
      "\n",
      "from __future__ import division\n",
      "\n",
      "\n",
      "b = 3                            # base of the channel\n",
      "\n",
      "z = 0.5                          # slope of the channel\n",
      "\n",
      "y = 2                            # depth of the channel\n",
      "\n",
      "\n",
      "T = b + 2*z*y\n",
      "\n",
      "print \"Top width =\",round(T,0),\"m\"\n",
      "\n",
      "A = (b+z*y)*y\n",
      "\n",
      "print \"Area of flow =\",round(A,0),\"m**2\"\n",
      "\n",
      "P = b + 2*y*sqrt(1+z**2)\n",
      "\n",
      "print \"Wetted perimeter =\",round(P,3),\"m\"\n",
      "\n",
      "R = A/P\n",
      "\n",
      "print \"Hydraulic radius =\",round(R,2),\"m\"\n",
      "\n",
      "D = A/T\n",
      "\n",
      "print \"Hydraulic depth =\",round(D,2),\"m\"\n",
      "\n",
      "Z = A*sqrt(D)\n",
      "\n",
      "print \"Secton Factor =\",round(Z,2),\"m**2\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Top width = 5.0 m\n",
        "Area of flow = 8.0 m**2\n",
        "Wetted perimeter = 7.472 m\n",
        "Hydraulic radius = 1.07 m\n",
        "Hydraulic depth = 1.6 m\n",
        "Secton Factor = 10.12 m**2\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 10.2 Page no 366"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import *\n",
      "\n",
      "from __future__ import division\n",
      "\n",
      "\n",
      "z = 1.0                # slide slope\n",
      "\n",
      "b = 3.0                # base width\n",
      "\n",
      "y = 1.5                # depth\n",
      "\n",
      "S = 0.0009\n",
      "\n",
      "n = 0.012             # for concrete\n",
      "\n",
      "\n",
      "A = (b+z*y)*y\n",
      "\n",
      "P = P = b + 2*y*sqrt(1+z**2)\n",
      "\n",
      "R = A/P\n",
      "\n",
      "\n",
      "Q = A*(1/n)*(R**(2/3)*S**(1/2))\n",
      "\n",
      "print \"Discharge for the channel =\",round(Q,2),\"m**3/s\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge for the channel = 16.1 m**3/s\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 10.4 Page no 373"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from __future__ import division\n",
      "\n",
      "from math import *\n",
      "\n",
      "\n",
      "z = 1\n",
      "\n",
      "Q = 10000/60                     # discharge of water in ft**#/s\n",
      "\n",
      "\n",
      "y = (Q/(1.828*2.25*sqrt(0.5)))**(2/5)\n",
      "\n",
      "print \"depth(y) =\",round(y,2),\"ft\"\n",
      "\n",
      "b = 0.828*y\n",
      "\n",
      "print \"base width(b) =\",round(b,2),\"ft\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "depth(y) = 5.05 ft\n",
        "base width(b) = 4.18 ft\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 10.5 Page no 378"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import *\n",
      "\n",
      "from __future__ import division\n",
      "\n",
      "\n",
      "y = 2.5                 # depth\n",
      "\n",
      "V = 8                  # velocity in m/s\n",
      "\n",
      "g = 9.81               # acceleration due to gravity in m/s**2\n",
      "\n",
      "\n",
      "Yc = (20**2/g)**(1/3)\n",
      "\n",
      "print \"Critical depth =\",round(Yc,2),\"m\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Critical depth = 3.44 m\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 10.6 Page no 380"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import *\n",
      "\n",
      "from __future__ import division\n",
      "\n",
      "\n",
      "Q = 15 # flow rate in m**3/s\n",
      "\n",
      "w = 4.0  # bottom width\n",
      "\n",
      "S = 0.0008 # bed slope\n",
      "\n",
      "n = 0.025 # manning constant\n",
      "\n",
      "z = 0.5 # slope\n",
      "\n",
      "\n",
      "\n",
      "y = 2.22 # we take the value of y as 2.2 m\n",
      "\n",
      "Q = ((4+0.5*y)*(y/(n))*(((4+0.5*y)*y)/(4+2.236*y))**(0.667)*(S)**(0.5))\n",
      "\n",
      "print \"a )Normal Depth =\",round(y,2),\"m\"\n",
      "\n",
      "A = (4+0.5*y)*y\n",
      "\n",
      "T = (w+2*z*y)\n",
      "\n",
      "D = A/T\n",
      "\n",
      "V = (Q/A)\n",
      "\n",
      "F =V/(sqrt(9.81*D)) \n",
      "\n",
      "print \"b )F = \",round(F,2),\" Since the Froude number is less than 1, the flow is subcritical\"\n",
      "\n",
      "\n",
      "yc = 1.08\n",
      "\n",
      "Q1 = (4+z*yc)*yc*sqrt((9.81*(4+0.5*yc)*yc)/(4+2*z*yc)) \n",
      "\n",
      "print \"c )Critical depth is = \",round(yc,2),\"m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a )Normal Depth = 2.22 m\n",
        "b )F =  0.31  Since the Froude number is less than 1, the flow is subcritical\n",
        "c )Critical depth is =  1.08 m\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "\n",
      "Example 10.8 Page no 390"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import *\n",
      "\n",
      "from __future__ import division\n",
      "\n",
      "\n",
      "b = 60                   # base width in ft\n",
      "\n",
      "y1 = 2.5                 # base depth in ft\n",
      "\n",
      "Q = 2500                 # discharge in ft**3/s\n",
      "\n",
      "g = 32.2\n",
      "\n",
      "\n",
      "V1 = Q/(b*y1)\n",
      "\n",
      "F1 = V1/sqrt(g*y1)\n",
      "\n",
      "y2 = y1*0.5*(sqrt(1+8*F1**2)-1)\n",
      "\n",
      "V2 = Q/(b*y2)\n",
      "\n",
      "print \"Since F1 =\",round(F1,2),\" It is a weak jump\"\n",
      "\n",
      "L = y2*4.25\n",
      "\n",
      "print \"Length of the jump =\",round(L,0),\"ft\"\n",
      "\n",
      "E1 = y1+(V1**2/(2*g))\n",
      "\n",
      "E2 = y2+(V2**2/(2*g))\n",
      "\n",
      "El = E1-E2\n",
      "\n",
      "Te = El*62.4*Q/543\n",
      "\n",
      "print \"Total energy loss =\",round(Te,2),\"HP\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Since F1 = 1.86  It is a weak jump\n",
        "Length of the jump = 23.0 ft\n",
        "Total energy loss = 133.7 HP\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 10.12 Page no 409"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from math import *\n",
      "\n",
      "from __future__ import division\n",
      "\n",
      "\n",
      "d = 6                      # depth of the channel\n",
      "\n",
      "w= 12                      # width of the channel\n",
      "\n",
      "h  = 1.0                   # height of the channel\n",
      "\n",
      "p = 9                      # pressure drop in m\n",
      "\n",
      "g = 32.2\n",
      "\n",
      "\n",
      "y2 = y1 - h - 0.75\n",
      "\n",
      "V1 = sqrt(2*g*0.75/((1.41)**2-1))\n",
      "\n",
      "Q = w*b*V1/10\n",
      "\n",
      "print \"Discharge =\",round(Q,0),\"cfs\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge = 503.0 cfs\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}