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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3 :  Flow Through Orifices Mouthpieces Nozzles"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.1 page no : 70"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#initialisation of variables\n",
      "\n",
      "import math \n",
      "Q = 16. \t\t\t#gpm\n",
      "w = 62.4 \t\t\t#lb/ft**3\n",
      "d = 1. \t\t\t#in\n",
      "h = 2+(5./12) \t\t\t#ft\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "x = 11.5 \t\t\t#ft\n",
      "h1 = 1.2 \t\t\t#in\n",
      "#CALCULATIONS\n",
      "Cd = Q*10/(60*w*(math.pi*(d/12)**2/4)*math.sqrt(2*g*h))\n",
      "Cv = math.sqrt(x**2/(4*(h1/12)*h*12**2))\n",
      "Cc = Cd/Cv\n",
      "Cr = (1-Cv**2)/Cv**2\n",
      "#RESULTS\n",
      "print  'Cc  = %.3f  '%(Cc)\n",
      "print  'Cv  = %.3f  '%(Cv)\n",
      "print  'Cd  = %.3f  '%(Cd)\n",
      "print  'Cr  = %.3f  '%(Cr)\n",
      "\n",
      "# note : answers are slightly different because of rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Cc  = 0.644  \n",
        "Cv  = 0.975  \n",
        "Cd  = 0.628  \n",
        "Cr  = 0.053  \n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.2 page no : 71"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation of variables\n",
      "Ww = 261. \t\t\t#lb/min\n",
      "a = 1. \t\t\t#in**2\n",
      "h = 4. \t\t\t#ft\n",
      "y =  5. \t\t\t#ft\n",
      "W1 = 10.65 \t\t\t#lb\n",
      "l = 1. \t\t\t#ft\n",
      "Q = 261. \t\t\t#lb/min\n",
      "w = 62.4 \t\t\t#lb/ft**3\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "#CALCULATIONS\n",
      "v = Q*144./(w*60)\n",
      "F = W1*l/y\n",
      "v = F*g*60./Q\n",
      "vth = math.sqrt(2*g*h)\n",
      "Cv = v/vth\n",
      "Q1 = Ww/w\n",
      "Qth = vth*60./144\n",
      "Cd = Q1/Qth\n",
      "Cc = Cd/Cv\n",
      "#RESULTS\n",
      "print  'Cv  = %.3f  '%(Cv)\n",
      "print  'Cd  = %.3f  '%(Cd)\n",
      "print  'Cc  = %.3f  '%(Cc)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Cv  = 0.982  \n",
        "Cd  = 0.625  \n",
        "Cc  = 0.637  \n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.3 page no : 74"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#initialisation of variables\n",
      "\n",
      "import math \n",
      "Q = 10. \t\t\t#ft**3/sec\n",
      "a1 = 1. \t\t\t#ft**2\n",
      "a2 = 4. \t\t\t#ft**2\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "p1 = 12. \t\t\t#lb/in**2\n",
      "v1 = 10. \t\t\t#ft/sec\n",
      "w = 62.4 \t\t\t#lb/ft**3\n",
      "#RESULTS\n",
      "v2 = v1*a1/a2\n",
      "Hl = (v1-v2)**2/(2*g)\n",
      "p2 = ((p1*144/w)+(v1**2/(2*g))-(v2**2/(2*g))-Hl)*(w/144)\n",
      "W = Hl*v1*w/550.\n",
      "\n",
      "#RESULTS\n",
      "print  'Head lost  = %.3f ft of water '%(Hl)\n",
      "print  'Pressure in larger part of pipe  = %.2f lb/in**2 '%(p2)\n",
      "print  'Work done  = %.3f HP '%(W)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Head lost  = 0.873 ft of water \n",
        "Pressure in larger part of pipe  = 12.25 lb/in**2 \n",
        "Work done  = 0.991 HP \n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.4 page no : 78"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#initialisation of variables\n",
      "\n",
      "import math \n",
      "Cc = 1.\n",
      "Cv = 0.833\n",
      "d = 2. \t\t\t#in\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "H = 12. \t\t\t#ft\n",
      "Pa = 34. \t\t\t#lb/in**2\n",
      "#/CALCULATIONS\n",
      "Q = Cc*Cv*math.pi*(d/12)**2*math.sqrt(2*g*H)/4\n",
      "Cd = Cc*Cv\n",
      "Pc = Pa-0.92*H\n",
      "#RESULTS\n",
      "#RESULTS\n",
      "print  'Discharge  = %.3f cu ft/sec '%(Q)\n",
      "print  'Coefficient of discharge  = %.3f  '%(Cd)\n",
      "print  'Pressure at Vent-contraction = %.2f ft of water '%(Pc)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge  = 0.505 cu ft/sec \n",
        "Coefficient of discharge  = 0.833  \n",
        "Pressure at Vent-contraction = 22.96 ft of water \n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.5 page no : 81"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "H = 4. \t\t\t#ft\n",
      "d = 1. \t\t\t#in\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "Cc = 0.5\n",
      "#CALCULATIONS\n",
      "Q = Cc*math.pi*(d/12)**2*math.sqrt(2*g*H)/4\n",
      "#RESULTS\n",
      "print  'Actual Discharge  = %.4f cu ft/sec '%(Q)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Actual Discharge  = 0.0438 cu ft/sec \n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.6 page no : 82"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation of variables\n",
      "\n",
      "D = 4. \t\t\t#ft\n",
      "d = 2. \t\t\t#in\n",
      "H1 = 6. \t\t\t#ft\n",
      "H2 = 2. \t\t\t#ft\n",
      "t = 4. \t\t\t#min\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "w = 62.4 \t\t\t#lb/ft**3\n",
      "H = 5. \t\t\t#ft\n",
      "#CALCULATIONS\n",
      "Cd = (2.*(math.pi/4.)*D**2*(math.sqrt(H1)-math.sqrt(H2)))/(t*60*(math.pi/4)*(d/12)**2*math.sqrt(2*g))\n",
      "Q = Cd*(math.pi/4)*(d/12)**2*math.sqrt(2*g*H)*w*60/10\n",
      "#RESULTS\n",
      "print  'Cd  = %.3f '%(Cd)\n",
      "print  'Discharge  = %.1f gpm'%(Q)\n",
      "\n",
      "# note : answers are slightly different because of rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Cd  = 0.619 \n",
        "Discharge  = 90.8 gpm\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.7 page no : 84"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation of variables\n",
      "H1 = 10. \t\t\t#ft\n",
      "H2 = 2. \t\t\t#ft\n",
      "Cd = 0.61\n",
      "d1 = 8. \t\t\t#ft\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "d2 = 3. \t\t\t#ft\n",
      "#CALCULATIONS\n",
      "a = d2**2./144\n",
      "H0 = H1*d2/(d1-d2)\n",
      "t = math.pi*(d1/2)**2*((2/5.)*(H1**(5./2)-H2**(5./2))+2*H0**2*(math.sqrt(H1)- \\\n",
      "math.sqrt(H2))+(4./3)*H0*(H1**(3./2)-H2**(3./2)))/(60*Cd*a*math.sqrt(2*g)*(H1+H0)**2)\n",
      "\n",
      "#RESULTS\n",
      "print  'time required to lower the water level  = %.2f min'%(t)\n",
      "\n",
      "# Note : answer is different because of rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "time required to lower the water level  = 5.14 min\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.8 page no : 85"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "D = 10. \t\t\t#ft\n",
      "H1 = 17. \t\t\t#ft\n",
      "H2 = 5. \t\t\t#ft\n",
      "d = 3. \t\t\t#in\n",
      "Cd = 0.62\n",
      "g =32.2 \t\t\t#ft/s**2\n",
      "#CALCULATIONS\n",
      "t1 = (2*math.pi*D**2/4)*(math.sqrt(H1)-math.sqrt(H2))/(Cd*math.sqrt(2*g)*math.pi*(d/12)**2/4)\n",
      "t2 = math.pi*(14./15)*H2**(5./2)*4/(Cd*math.pi*(d/12.)**2*math.sqrt(2*g))\n",
      "t = t1+t2\n",
      "#RESULTS\n",
      "print  'time required to empty the vessel  = %.f sec'%(t)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "time required to empty the vessel  = 1885 sec\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.9 page no : 86"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation of variables\n",
      "Cd = 0.8\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "d = 3. \t\t\t#in\n",
      "#CALCULATIONS\n",
      "t = (60*2/(math.pi*(d/12)**2*math.sqrt(2*g)/4*Cd))*(6-d)**(3./2)/(3*60./2)\n",
      "#RESULTS\n",
      "print  'time to emptify biler  = %.2f min'%(t)\n",
      "\n",
      "# note : answer is different because of rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "time to emptify biler  = 21.98 min\n"
       ]
      }
     ],
     "prompt_number": 33
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.10 page no : 87"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "#initialisation of variables\n",
      "A = 100 * 27      # sq ft\n",
      "dif = 8 - 3.      # ft\n",
      "a = 2.            # sq ft\n",
      "Cd = 0.8          # Co-efficient\n",
      "\n",
      "#CALCULATIONS\n",
      "t1 = round(A*((a/3 * 22.7) - a/3 * 5.19 - (a/3*11.2))/(Cd*2*8.02*dif))\n",
      "t2 = round(A*(2./3)*11.2/(Cd*a*8.02*dif))\n",
      "t = t1 + t2\n",
      "\n",
      "#RESULTS\n",
      "print  'Total time to empty the tank = %d sec'%t"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total time to empty the tank = 491 sec\n"
       ]
      }
     ],
     "prompt_number": 45
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.11 page no : 89"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "H1 = 9. \t\t\t#ft\n",
      "H2 = 4. \t\t\t#ft\n",
      "Cd = 0.6\n",
      "a = 4. \t\t\t#in**2\n",
      "A1 = 72. \t\t\t#ft**2\n",
      "A2 = 24. \t\t\t#ft**2\n",
      "g =32.2 \t\t\t#ft/s**2\n",
      "#CALCULATIONS\n",
      "t = (2*A1*A2/(A1+A2))*(math.sqrt(H1)-math.sqrt(H2))*144/(Cd*60*a*math.sqrt(2*g))\n",
      "#RESULTS\n",
      "print  'time required to reduce the water level difference  = %.1f min'%(t)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "time required to reduce the water level difference  = 4.5 min\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.12 pageno : 91"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#initialisation of variables\n",
      "import math \n",
      "l = 80. \t\t\t#ft\n",
      "w = 12. \t\t\t#ft\n",
      "t = 3. \t\t\t#min\n",
      "Hl = 12. \t\t\t#ft\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "Cd = 0.6\n",
      "#CALCULATIONS\n",
      "s = math.sqrt(2*l*w*Hl**(1./2)/(Cd*math.sqrt(2*g)*t*60.))\n",
      "#RESULTS\n",
      "print  'side of the square orifice  = %.2f ft'%(s)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "side of the square orifice  = 2.77 ft\n"
       ]
      }
     ],
     "prompt_number": 46
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.13 page no : 92"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#initialisation of variables\n",
      "import math \n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "Cd = 0.6\n",
      "d = 2. \t\t\t#in\n",
      "H1 = 5. \t\t\t#ft\n",
      "\n",
      "#CALCULATIONS\n",
      "K = round(Cd * math.pi/4 * (d/12)**2 * math.sqrt(g*2),3)\n",
      "t = d*math.pi*(0.5*math.log(1.89) - 0.235)/K**2 \n",
      "v = round(math.sqrt(2*g*H1)/2.)\n",
      "q = v*Cd*math.pi*(d/12)**2./4\n",
      "\n",
      "#RESULTS\n",
      "print \"Time required to raise the level is : %.2f sec\"%t\n",
      "print  'Total discharge  = %.3f cfs'%(q)\n",
      "\n",
      "# Note : answers may vary because of rounding error. Please calculate manually."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time required to raise the level is : 47.47 sec\n",
        "Total discharge  = 0.118 cfs\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.14 page no : 95"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "Cd = 0.62\n",
      "H = 9. \t\t\t#in\n",
      "l = 3. \t\t\t#ft\n",
      "g = 32.2 \t\t\t#t/sec**2\n",
      "#CALCULATIONS\n",
      "Q1 = Cd*(H*l/12)*math.sqrt(2*g*3*H/24.)\n",
      "Q2 = Cd*2*l*math.sqrt(2*g)*((H/6)**(3./2)-(H/12)**(3./2))/3\n",
      "#RESULTS\n",
      "print  'Discharge by appropriate formula  = %.2f cfs'%(Q1)\n",
      "print  ' Discharge by exact formula  = %.2f cfs'%(Q2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Discharge by appropriate formula  = 11.87 cfs\n",
        " Discharge by exact formula  = 11.82 cfs\n"
       ]
      }
     ],
     "prompt_number": 54
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 3.15 pageno : 95"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "#initialisation of variables\n",
      "\n",
      "Cd = 0.62\n",
      "B = 2.5 \t\t\t#ft\n",
      "H2 = 8. \t\t\t#ft\n",
      "H1 = 7. \t\t\t#ft\n",
      "g = 32.2 \t\t\t#ft/sec**2\n",
      "h = 4. \t\t\t#ft\n",
      "#CALCULATIONS\n",
      "Q1 = round(2*Cd*B*math.sqrt(2*g)*(H2**(3./2)-H1**(3./2))/3)\n",
      "Q2 = Cd*math.sqrt(2*g)*math.sqrt(H2)*B*(h-1)\n",
      "Q = Q1+Q2\n",
      "#RESULTS\n",
      "print  'Total discharge  = %d cfs'%(Q)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total discharge  = 139 cfs\n"
       ]
      }
     ],
     "prompt_number": 60
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}