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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 2 : Fluid statics"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.1 page no :  40\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# variables\n",
      "g=32.2;                     #ft/s^2\n",
      "rho_water=62.3;             #lbm/ft^3\n",
      "\n",
      "# calculation\n",
      "#specific weoight=(density)*(acceleration due to gravity)\n",
      "specific_wt=rho_water*g;    #lbm.ft/ft^3.s^2\n",
      "\n",
      "#1 lbf=32.2 lbm.ft/s^2\n",
      "specific_wt=specific_wt/32.2;            #lbf/ft^3\n",
      "\n",
      "# result\n",
      "print \"Specific weight of water is\" ,specific_wt , \"lbf/ft^3\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Specific weight of water is 62.3 lbf/ft^3\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.2 page no : 40\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# variables\n",
      "d=304.9;                       #m\n",
      "rho_water=1024.;               #Kg/m^3\n",
      "g=9.81;                        #m/s^2\n",
      "p_atm=101.3;                   #KPa\n",
      "\n",
      "# calculation\n",
      "#gauge pressure=(desity)*(acc. due to gravity)*(depth)\n",
      "p_depth=p_atm+rho_water*g*d/1000.0;            #KPa\n",
      "\n",
      "# result\n",
      "print \"pressure at the depth is\" , (p_depth) , \"KPa\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "pressure at the depth is 3164.154656 KPa\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.3 page no : 41\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# variables\n",
      "rho_oil=55.;                        #lbm/ft^3\n",
      "g=32.2;                             #ft/s^2\n",
      "d=60.;                              #ft (depth of oil cylinder)\n",
      "\n",
      "# calculation and result\n",
      "gauge_pressure=rho_oil*g*d/32.2;    #lbf/ft^2\n",
      "print \"Gauge pressure is\",\n",
      "print gauge_pressure,\n",
      "print \"lbf/ft^2\"\n",
      "\n",
      "#1 ft=12 in\n",
      "gauge_pressure=gauge_pressure/144.0;        #lbf/in^2\n",
      "print \"Gauge pressure is\",\n",
      "print gauge_pressure,\n",
      "print \"lbf/in^2\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Gauge pressure is 3300.0 lbf/ft^2\n",
        "Gauge pressure is 22.9166666667 lbf/in^2\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.4 page no : 42\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "\n",
      "# varirbles\n",
      "#calc of density of air at a certain height\n",
      "p_atm=14.7;                          #psia\n",
      "T=289.;                              #K\n",
      "\n",
      "#P2=P1*exp^(-(acc. due to gravity)*(mass of air)*(height)/(universal gas const.)/(temp.))\n",
      "g=9.81;                              #m/s^2\n",
      "R=8314;                              #N.m^2/Kmol/K\n",
      "\n",
      "#for height of 1000 ft=304.8m\n",
      "h=304.8;                             #m\n",
      "p_1000=14.7*math.exp(-g*29*h/R/289);\n",
      "print \"pressure at 1000ft is\",\n",
      "print p_1000,\n",
      "print \"psia\"\n",
      "\n",
      "#for height of 10000 ft=3048m\n",
      "h=3048.;                            #m\n",
      "p_10000=p_atm*math.exp(-g*29.*h/R/289.);\n",
      "print \"pressure at 10000ft is\",\n",
      "print p_10000,\n",
      "print \"psia\"\n",
      "\n",
      "#for height of 100000 ft=30480m\n",
      "h=30480.;                           #m\n",
      "p_100000=14.7*math.exp(-g*29.*h/R/289.);\n",
      "print \"pressure at 100000ft is\",\n",
      "print p_100000,\n",
      "print \"psia\","
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "pressure at 1000ft is 14.1789512072 psia\n",
        "pressure at 10000ft is 10.2467246829 psia\n",
        "pressure at 100000ft is 0.398102276652 psia\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.5 page no : 42\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# variables\n",
      "p_atm=14.7;                        #psia\n",
      "g=9.81;                            #m/s^2\n",
      "\n",
      "#P2=P1*[1-(acc. due to gravity)*(mass of air)*(height)/(univ. gas const.)/(temp.)]\n",
      "T=289.;                            #K\n",
      "R=8314.                            #N.m^2/Kmol/K\n",
      "\n",
      "\n",
      "# calculation and result\n",
      "#for height of 1000ft=304.8m\n",
      "h=304.8                            #m\n",
      "p_1000=p_atm*(1-g*29*h/R/T)\n",
      "print \"pressure at 1000ft is\",\n",
      "print p_1000,\n",
      "print \"psia\"\n",
      "\n",
      "#for height of 10000ft=3048m\n",
      "h=3048.                            #m\n",
      "p_10000=p_atm*(1-g*29*h/R/T)\n",
      "print \"pressure at 10000ft is\",\n",
      "print p_10000,\n",
      "print \"psia\"\n",
      "\n",
      "#for height of 100000ft=30480m\n",
      "h=30480.                           #m\n",
      "p_100000=p_atm*(1-g*29*h/R/T)\n",
      "print \"pressure at 100000ft is\",\n",
      "print p_100000,\n",
      "print \"psia\"\n",
      "\n",
      "#NOTE that the pressure comes out to be negative at 100000ft justifying that density of air changes with altitude"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "pressure at 1000ft is 14.1694926079 psia\n",
        "pressure at 10000ft is 9.39492607874 psia\n",
        "pressure at 100000ft is -38.3507392126 psia\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.6 page no : 45\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "\n",
      "# variables\n",
      "#calc atm pressure on a storage tank roof\n",
      "p_atm=14.7;                    #psia\n",
      "\n",
      "#diameter of roof is 120ft\n",
      "d_roof=120.;                   #ft\n",
      "\n",
      "# calculation\n",
      "#force=(pressure)*(area)\n",
      "f_roof=p_atm*(math.pi)*d_roof**2/4.*144;            #lbf ;144 because 1ft=12inch\n",
      "\n",
      "# result\n",
      "print \"Force exerted by atmosphere on the roof is\",\n",
      "print f_roof,\n",
      "print \"lbf\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Force exerted by atmosphere on the roof is 23940443.9848 lbf\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.7 page no : 45\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "\n",
      "# variables\n",
      "#calc atm pressure on a storage tank roof\n",
      "p_atm=14.7;                             #psia\n",
      "\n",
      "#diameter of roof is 120ft\n",
      "d_roof=120.;                            #ft\n",
      "#force=(atm. pressure + gauge pressure)*(area)\n",
      "#gauge pressure=(desity)*(acc. due to gravity)*(depth)\n",
      "rho_water=62.3                          #lbm/ft^3\n",
      "g=32.2;                                 #ft/s^2\n",
      "\n",
      "# calculation\n",
      "#depth of water on roof=8 inch=o.667 ft\n",
      "h=0.667;                                #ft\n",
      "gauge_pressure=rho_water*g*h/32.2*(math.pi)*d_roof**2/4.;                 #lbf\n",
      "\n",
      "# result\n",
      "print gauge_pressure"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "469965.799032\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.8 page no : 46\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# variables\n",
      "#lock gate has water on one side and air on the other at atm. pressure\n",
      "w=20.;                       #m (width of the lock gate)\n",
      "h=10.;                       #m (height of the lock gate)\n",
      "p_atm=1.;                    #atm\n",
      "rho_water=1000.;             #Kg/m^3\n",
      "g=9.81                       #m/s^2\n",
      "\n",
      "# calculation\n",
      "#for a small strip of dx height at the depth of x on the lock gate\n",
      "#net pressure on strip = (p_atm+(rho_water)*g*x) - p_atm\n",
      "#thus, net pressure on strip = (rho_water)*g*x\n",
      "#force on strip = (rho_water*g*x)*w.dx = (rho_water)*g*w*(x.dx)\n",
      "#force on lock gate = integration of force on strip fromm h=0 to h=10\n",
      "#integration(x.dx) = x^2/2\n",
      "#for h=0 to h=10; integration (x.dx) = h^2/2\n",
      "force_lockgate=(rho_water)*g*w*h**2/2;\n",
      "\n",
      "# result\n",
      "print \"The net force on the lock gate is\",force_lockgate/10**6,\"MN\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The net force on the lock gate is 9.81 MN\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "example 2.9 page no : 49\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "sigma_tensile=20000.                        #lbf/in^2 (tensile stress is normally 1/4 rupture stress)\n",
      "\n",
      "#max pressure is observed at the bottom of the storage\n",
      "p_max=22.9;                                 #lbf/in^2\n",
      "\n",
      "#diameter of storaeg tank = 120ft =1440in\n",
      "d=1440.;                                    #in\n",
      "\n",
      "# calculation\n",
      "t=(p_max)*d/sigma_tensile/2;                #in\n",
      "\n",
      "# result\n",
      "print \"Thichness of the storage tank is\",\n",
      "print t,\n",
      "print \"in\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Thichness of the storage tank is 0.8244 in\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.10 page no : 50\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# variables\n",
      "p_working=250.0;                          #lbf/in^2\n",
      "\n",
      "#diameter of the cylinder = 10ft = 120in\n",
      "d=120.0;                                   #in\n",
      "sigma_tensile=20000.;                      #lbf/in^2\n",
      "\n",
      "# calculation\n",
      "t=p_working*d/sigma_tensile/2;             #in\n",
      "\n",
      "# result\n",
      "print \"Thichness of the storage tank is\",\n",
      "print t,\n",
      "print \"in\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Thichness of the storage tank is 0.75 in\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.11 page no :  53\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "\n",
      "# variables\n",
      "p_atm=1.;                      #atm\n",
      "T=293.;                        #K\n",
      "d=3.;                          #m (diameter of the balloon)\n",
      "\n",
      "# calculation\n",
      "#buoyant force=(density of air)*g*(volume of balloon)\n",
      "#weight of balloon = (density of helium)*g*(volume of balloon)\n",
      "#density for gases = PM/RT\n",
      "#payload of balloon = buoyant force - weight\n",
      "V_balloon=(math.pi)*d**3/6.;      #m^3\n",
      "R=8.2*10**(-2);                   #m^3.atm/mol/K\n",
      "M_air=29.;                        #Kg/Kmol\n",
      "M_he=4.;                          #Kg/Kmol\n",
      "g=9.81;                           #m/s^2\n",
      "payload=(V_balloon)*g*p_atm*(M_air-M_he)/R/T;            #N\n",
      "\n",
      "# result\n",
      "print \"Payload of the balloon is\",\n",
      "print payload,\n",
      "print \"N\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Payload of the balloon is 144.307841185 N\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.12 page no : 54\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# variables\n",
      "#calc fraction of block in water\n",
      "SG_wood=0.96;                  #Specific gravity\n",
      "SG_gasoline=0.72;\n",
      "\n",
      "# calculation\n",
      "#Let r be the ratio - V_water/V_wood\n",
      "r=(SG_wood-SG_gasoline)/(1-SG_gasoline);\n",
      "\n",
      "# result\n",
      "print \"Fraction of wood in water\",\n",
      "print r"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Fraction of wood in water 0.857142857143\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.13 page no : 54\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# variables\n",
      "#height of water above pt.C = 2.5ft\n",
      "rho_water=62.3;                 #lbm/ft^3;\n",
      "h1=2.5;                         #ft\n",
      "rho_gas=0.1;                    #lbm/ft^3\n",
      "h2=0.5;                         #ft (height of gas)\n",
      "g=32.2;                         #ft/s^2\n",
      "\n",
      "# calculation\n",
      "gauge_pressure=((rho_water)*g*h1+(rho_gas)*g*h2)/144/32.2          #lbf/in^2\n",
      "\n",
      "# result\n",
      "print \"Gauge pressure is\",\n",
      "print gauge_pressure,\n",
      "print \"lbf/in^2\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Gauge pressure is 1.08194444444 lbf/in^2\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.14 page no : 56\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "rho_water=62.3;                       #lbm/ft^3\n",
      "SG_oil=1.1;\n",
      "rho_oil=SG_oil*(rho_water);\n",
      "g=32.2;                               #ft/s^2\n",
      "h1_1=1.;                              #ft\n",
      "h1_2=2.;                              #ft\n",
      "h2_1=2.;                              #ft\n",
      "h2_2=1.;                              #ft\n",
      "\n",
      "# calculation\n",
      "p_diff=((rho_water)*g*(h1_1-h1_2)+(rho_oil)*g*(h2_1-h2_2))/32.2/144.0;        #lbf/in^2\n",
      "\n",
      "# result\n",
      "print \"The pressure difference is\",\n",
      "print p_diff,\n",
      "print \"lbf/in^2\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The pressure difference is 0.0432638888889 lbf/in^2\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.15 page no :  57\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# variables\n",
      "k=10000.;                      #N/m (spring constant)\n",
      "x=0.025;                       #m (displacement in spring)\n",
      "A=0.01;                        #m^2 (area of piston)\n",
      "\n",
      "# calculation\n",
      "gauge_pressure=k*x/A/1000.;    #KPa\n",
      "\n",
      "# result\n",
      "print \"The gauge pressure is\",\n",
      "print gauge_pressure,\n",
      "print \"KPa\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The gauge pressure is 25.0 KPa\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.16 page no : 60\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# variables\n",
      "g=32.2;                          #ft/s^2\n",
      "h=20.;                           #ft (height of fireplace)\n",
      "rho_air=0.075;                   #lbm/ft^3\n",
      "T_air=293.0;                     #K (surrounding temperature)\n",
      "T_fluegas=422.0;                 #K\n",
      "\n",
      "# calculation\n",
      "p_diff=g*h*(rho_air)*(1-(T_air/T_fluegas))/32.2/144;             #lbf/in^2\n",
      "\n",
      "# result\n",
      "print \"The pressure difference is\",\n",
      "print p_diff,\n",
      "print \"lbf/in^2\","
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The pressure difference is 0.00318424170616 lbf/in^2\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.17 page no : 64\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# variables\n",
      "rho_water=1000.                       #Kg/m^3\n",
      "g=9.81;                               #m/s^2\n",
      "h=5.;                                 #m (depth of water)\n",
      "\n",
      "# calculation and result\n",
      "#for elevator not accelerated\n",
      "p_gauge=(rho_water)*g*h/1000.0;       #KPa\n",
      "print \"THe gauge pressure is\",\n",
      "print p_gauge,\n",
      "print \"KPa\"\n",
      "\n",
      "#for elevator accelerated at 5m/s^2 in upward direction\n",
      "a=5.;                                 #m/s^2\n",
      "p_gauge=(rho_water)*(g+a)*h/1000.0;   #KPa\n",
      "print \"THe gauge pressure is\",\n",
      "print p_gauge,\n",
      "print \"KPa\"\n",
      "\n",
      "#for elevator accelerated at 5m/s^2 in downward direction\n",
      "a=5.;                                 #m/s^2\n",
      "p_gauge=(rho_water)*(g-a)*h/1000.0;   #KPa\n",
      "print \"THe gauge pressure is\",\n",
      "print p_gauge,\n",
      "print \"KPa\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "THe gauge pressure is 49.05 KPa\n",
        "THe gauge pressure is 74.05 KPa\n",
        "THe gauge pressure is 24.05 KPa\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "example 2.18 page no : 65\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "\n",
      "# variables\n",
      "#angle free surface makes with the horizontal in an accelerated body\n",
      "a=1.;                             #ft/s^2\n",
      "g=32.2;                           #ft/s^2\n",
      "\n",
      "# calculation\n",
      "theta=math.atan(a/g);             #radians\n",
      "theta=theta*180./math.pi;         #degrees\n",
      "\n",
      "# result\n",
      "print \"The angle made by free surface with the horizontal is\",\n",
      "print theta,\n",
      "print \"degrees\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The angle made by free surface with the horizontal is 1.77880031567 degrees\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.19 page no : 66\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "# variables\n",
      "f=78/60.0;                      #rps\n",
      "r=0.15;                         #m\n",
      "g=9.81;                         #m/s^2\n",
      "\n",
      "# calculation\n",
      "#omega=2*(%pi)*f\n",
      "z=((2*(math.pi)*f)**2)*r**2/2/g;             #m\n",
      "\n",
      "# result\n",
      "print \"The liquid in the cylinder rises to a height of\",\n",
      "print z,\n",
      "print \"m\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The liquid in the cylinder rises to a height of 0.0765120708158 m\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " example 2.20 page no : 67\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "\n",
      "# variables\n",
      "#Let difference between heights at bottom and top be d\n",
      "d=20.;                        #in\n",
      "r_a=14.;                      #in\n",
      "f=1000/60.;                   #rps\n",
      "g=32.2;                       #ft/s^2\n",
      "\n",
      "# calculation\n",
      "r_b=((r_a)**2-2*(d)*g*12/(2*(math.pi)*f)**2)**0.5;              #in\n",
      "\n",
      "# result\n",
      "print \"The thickness of water strip at bottom of industrial centrifuge\",\n",
      "print r_b,\n",
      "print \"in\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The thickness of water strip at bottom of industrial centrifuge 13.9495728181 in\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}