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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 13:Open-Channel Flow"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-1,Page No:711"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "V_dot=0.2 #Volumetric flow rate in m^3/s\n",
      "y=0.15 #Depth of flow in m\n",
      "b=0.4 #Width in m\n",
      "g=9.81 #Acceleration due to gravity in m/s^2\n",
      "\n",
      "#Calculations\n",
      "V=V_dot/(y*b) #Velocity in m/s\n",
      "yc=(V_dot**2/(g*b**2))**0.33 #Critical depth in m\n",
      "\n",
      "#Flow is supercritical\n",
      "\n",
      "Fr=V/((g*y)**0.5) #Froude Number\n",
      "\n",
      "Es1=y+(V_dot**2/(2*g*b**2*y**2)) #Specific Energy in m\n",
      "#Alternate Depth\n",
      "#Solving the Ploynomial Equation\n",
      "\n",
      "coeff=[1,-Es1,0,V_dot**2/(2*g*b**2)]\n",
      "x=numpy.roots(coeff)\n",
      "\n",
      "#Result\n",
      "print \"The velocity of flow is\",round(V,2),\"m/s\"\n",
      "print \"As the froude number Fr\",round(Fr,2),\"> 1 the flow is supercritical\"\n",
      "print \"The Alternate Depth is\",round(x[0],3),\"m\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The velocity of flow is 3.33 m/s\n",
        "As the froude number Fr 2.75 > 1 the flow is supercritical\n",
        "The Alternate Depth is 0.69 m\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-2, Page No:716"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "b=0.8 #Bottom width of the trapezoidal channel in m\n",
      "y=0.52 #Depth of flow in m\n",
      "theta=60 #Angle in degrees\n",
      "alpha1=0.3 #Slope angle in degrees\n",
      "n=0.03 #Mannings Coefficient \n",
      "a=1 #m^1/3/s\n",
      "alpha2=1 #Slope in degrees\n",
      "\n",
      "#Calculations\n",
      "Ac=y*(b+(y/tan((theta*pi)/180))) #Cross-sectional Area in m^2\n",
      "p=b+((2*y)/sin((theta*pi)/180)) #Perimeter in m\n",
      "Rh=Ac/p #Hydraulic Radius in m\n",
      "S01=tan((alpha1*pi)/180) #Slope of the bottom channel \n",
      "S02=tan((alpha2*pi)/180) #Slope of the bottom channel\n",
      "V_dot1=(a/n)*(Ac*Rh**0.66*S01**0.5) #Volumetric Flow rate in m^3/s\n",
      "V_dot2=(a/n)*(Ac*Rh**0.66*S02**0.5) #Volumetric Flow rate in m^3/s\n",
      "\n",
      "#Result\n",
      "print \"The volumetric flow rate when alpha is 0.3 degrees is\",round(V_dot1,2),\"m^3/s\" \n",
      "print \"The volumetric flow rate when alpha is 1 degrees is\",round(V_dot2,2),\"m^3/s\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The volumetric flow rate when alpha is 0.3 degrees is 0.6 m^3/s\n",
        "The volumetric flow rate when alpha is 1 degrees is 1.1 m^3/s\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-4, Page No:718"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "S0=0.003 #Bottom slope\n",
      "l1=3 #Length in m\n",
      "p1= 10.486 #Perimeter in section 1 in m\n",
      "p2=10 #Perimeter of section 2 in m\n",
      "Ac1=21 #Area of section 1 in m^2\n",
      "Ac2=16 #Area of section 2 in m^2\n",
      "a=1 #m^1/3/s\n",
      "n1=0.03 #Mannings coefficient for section 1\n",
      "n2=0.05 #Mannings Coefficient for section 2\n",
      "\n",
      "#Calcualtions\n",
      "Rh1=Ac1/p1 #Hydraulic Radius at section 1 in m\n",
      "Rh2=Ac2/p2 #Hydraulic Radius at section 2 in m\n",
      "Rh=(Ac1+Ac2)/(p1+p2) #Hydraulic Radius of the entire channel in m\n",
      "\n",
      "V_dot=a*S0**0.5*(((Ac1*Rh1**0.66)/n1)+((Ac2*Rh2**0.66)/n2)) #Volumetric Flow rate in m^3/s\n",
      "n_eff=(a*(Ac1+Ac2)*Rh**0.66*S0**0.5)/V_dot #Effective Mannings Coefficient\n",
      "\n",
      "#Result\n",
      "print \"The flow rate is\",round(V_dot),\"m^3/s\"\n",
      "print \"The effective Mannings Coefficient is\",round(n_eff,3)\n",
      "#The decimal point accuracy in python is the possible source of discrepancy in textbook and computed answer\n",
      "#The answer computed is weel within the permissbile error limit\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The flow rate is 78.0 m^3/s\n",
        "The effective Mannings Coefficient is 0.038\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-5, Page No:722"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#NOTE:Variable names have been changed\n",
      "\n",
      "#Variable Decleration\n",
      "n=0.016 #mannings Coefficient\n",
      "V_dot=2 #Volumetric Flow rate in m^3/s\n",
      "S0=0.001 #Bottom Slope\n",
      "theta=60 #Angle in degrees\n",
      "\n",
      "#Calculations\n",
      "\n",
      "#Part(a)\n",
      "#Using the Mannings Equation\n",
      "b1=((2*n*V_dot*4**0.66)/(a*S0**0.5))**0.375 #Width of the Rectangular section in m\n",
      "Aca=b1**2*0.5 #Area of rectangular section in m^2\n",
      "p=2*b1 #Perimeter in m\n",
      "y1=b1/2 #Depth of flow in m\n",
      "\n",
      "#Part(b)\n",
      "b2=((n*V_dot)/(0.75*3**0.5*((3**0.5/4)**0.66)*a*S0**0.5))**0.375 #Width in m\n",
      "y2=((3**0.5)/2)*b2 #Depth of flow in m\n",
      "\n",
      "#Result\n",
      "print \"The cross-section for rectangular section are b=\",round(b1,2),\"m y=\",round(y1,2),\"m\"\n",
      "print \"The cross-section for trapezoidal section are b=\",round(b2,2),\"m y=\",round(y2,3),\"m\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The cross-section for rectangular section are b= 1.84 m y= 0.92 m\n",
        "The cross-section for trapezoidal section are b= 1.12 m y= 0.97 m\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-7, Page No:732"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "y=2 #Depth of Flow in m\n",
      "b=6 #Bottom width in m\n",
      "a=1 #m^1/3/s\n",
      "S0=0.004 #Bed Slope\n",
      "n=0.014 #Mannings Coefficient\n",
      "g=9.81 #Acceleration due to gravity in m/s^2\n",
      "\n",
      "#Calculations\n",
      "Ac=y*b #Area in m^2\n",
      "p=b+2*y #Perimeter in m\n",
      "Rh=(Ac*10**-1)/(p*10**-1) #Hydraulic Radius in m \n",
      "\n",
      "#Flow rate \n",
      "V_dot=(a/n)*(Ac*Rh**0.66*S0**0.5) #Volumetric Flow rate in m^3/s\n",
      "\n",
      "#Critical Depth\n",
      "yc=V_dot**2/(g*Ac**2) #Critical Depth in m\n",
      "\n",
      "#Result\n",
      "print \"As yn=\",round(y,2),\"m < yc=\",round(yc,2),\"m the slope is STEEP\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "As yn= 2.0 m < yc= 2.65 m the slope is STEEP\n"
       ]
      }
     ],
     "prompt_number": 43
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-8,Page No:735"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "rho=1000 #Density of water in kg/m^3\n",
      "g=9.81 #Accelration due to gravity in m/s^2\n",
      "y1=0.8 #Pre jump height in m\n",
      "V1=7 #Velocity in m/s pre jump\n",
      "b=10 #Width of the channel in m\n",
      "\n",
      "#Calculations\n",
      "#Part(a)\n",
      "Fr1=V1/((g*y1)**0.5) #Froude Number pre Jump\n",
      "\n",
      "#Greater than 1 hence supecritical\n",
      "\n",
      "y2=0.5*y1*(-1+((1+8*Fr1**2)**0.5)) #Post Jump Height\n",
      "V2=(y1/y2)*V1 #Velocity post jump in m/s\n",
      "\n",
      "Fr2=V2/((g*y2)**0.5) #Froude Number after Jump\n",
      "\n",
      "#Part(b)\n",
      "h_L=y1-y2+(V1**2-V2**2)/(2*g) #Head Loss in m\n",
      "\n",
      "Es1=y1+V1**2/(2*g) #Specific Energy before jump in m\n",
      "\n",
      "Dissipation_Ratio=h_L/Es1 #Dissipiation Ratio\n",
      "\n",
      "#Part(c)\n",
      "m_dot=rho*b*y1*V1 #Mass Flow rate in kg/s\n",
      "\n",
      "E_dissipiated=m_dot*g*h_L #Energy Dissipiated in kW\n",
      "\n",
      "#Result \n",
      "print \"The Depth of flow after the Jump is\",round(y2,2),\"m and the Froude Number is\",round(Fr2,3)\n",
      "print \"The head loss is\",round(h_L,3),\"m and the Energy Dissipation Ratio is\",round(Dissipation_Ratio,3)\n",
      "print \"The energy wasted is\",round(E_dissipiated/1000),\"kW\"\n",
      "#NOTE:Answer differ due to decimal point accuracy"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "56000.0\n",
        "The Depth of flow after the Jump is 2.46 m and the Froude Number is 0.465\n",
        "The head loss is 0.577 m and the Energy Dissipation Ratio is 0.175\n",
        "The energy wasted is 317.0 kW\n"
       ]
      }
     ],
     "prompt_number": 57
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-9,Page No:738"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "y1=3 #Depth of flow in m\n",
      "a=0.25 #Height of Sluice Gate in m\n",
      "y2=1.5 #Depth of flow after the turbulence subsides in m\n",
      "Cd=0.47 #Coefficient of Discharge\n",
      "b=6 #Width of the channel in m\n",
      "g=9.81 #Acceleration due to gravity in m/s^2\n",
      "\n",
      "#Calculations\n",
      "depth_ratio1=y1/a #Depth ratio\n",
      "depth_ratio2=y2/a #Depth ratio\n",
      "V_dot=Cd*b*a*((2*g*y1)**0.5) #Volumetric Flow rate in m^3/s\n",
      "\n",
      "#Result\n",
      "print \"The volumetric Flow rate is\",round(V_dot,2),\"m^3/s\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The volumetric Flow rate is 5.41 m^3/s\n"
       ]
      }
     ],
     "prompt_number": 58
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-10, Page No:745"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "V1=1.2 #Velocity in m/s\n",
      "g=9.81 #Acceleration due to gravity in m/s^2\n",
      "y1=0.8 #Depth of flow before encounternign the bump in m\n",
      "delta_zb=0.15 # depth in m\n",
      "\n",
      "#Calculations\n",
      "Fr1=V1/((g*y1)**0.5) #Froude Number\n",
      "yc=((y1**2*V1**2)/g)**0.33 #Critical depth in m\n",
      "\n",
      "#Flow is subcritical\n",
      "Es1=y1+(V1**2/(2*g)) #Specific Energy in m\n",
      "\n",
      "#Solving the equation\n",
      "coeff=[1,-0.723,0,0.047]\n",
      "y=numpy.roots(coeff) #Depth of flow in m\n",
      "\n",
      "Depression=y1-(y[0]+delta_zb)\n",
      "\n",
      "#Result\n",
      "print \"The depression of the water surface is present and is\",round(Depression,2),\"m\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The depression of the water surface is present and is 0.06 m\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.13-11, Page No:746"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable Decleration\n",
      "y1=1.5 #Depth of flow in m\n",
      "Pw=0.6 #Height in m\n",
      "b=5 #Width in m\n",
      "g=9.81 #Acceleration due to gravity in m/s^2\n",
      "\n",
      "#Calculations\n",
      "H=y1-Pw #Weir Head in m\n",
      "\n",
      "#Using the Discharge Coefficient Formula\n",
      "Cwd_rec=0.598+(0.0897*(H/Pw)) #Coefficient of Discharge\n",
      "\n",
      "V_dot=(2*Cwd_rec*b*(2*g)**0.5*(H**1.5))/3 #Volumetric Flow rate in m^3/s\n",
      "\n",
      "#Result\n",
      "print \"The Volumetric Flow rate is\",round(V_dot,2),\"m^3/s\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Volumetric Flow rate is 9.23 m^3/s\n"
       ]
      }
     ],
     "prompt_number": 12
    }
   ],
   "metadata": {}
  }
 ]
}