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 "metadata": {
  "name": "",
  "signature": "sha256:eb1b6239947f111d6ea8f1bcaafe0bb84b121b2561bbb9f5f8a8bdec77169fd2"
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter2:SOME CONCEPTS AND DEFINITIONS"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2.1:pg-19"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#example 1\n",
      "#weight of a person\n",
      "\n",
      "\n",
      "m=1 #kg\n",
      "g=9.75 #acc.due to gravity in m/s^2\n",
      "F=m*g #weight of 1 kg mass in N\n",
      "print\"\\n hence,weight of person is\",round(F,2),\" N\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " hence,weight of person is 9.75  N\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2.2:pg-24"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#example 2\n",
      "#average volume and density\n",
      "\n",
      "Vliq=0.2 #volume of liquid in m^3\n",
      "dliq=997 #density of liquid in kg/m^3\n",
      "Vstone=0.12 #volume of stone in m^3\n",
      "Vsand=0.15 #volume of sand in m^3\n",
      "Vair=0.53 #vo;ume of air in m^3\n",
      "mliq=Vliq*dliq #mass of liquid in kg\n",
      "dstone=2750 #density of stone in kg/m^3\n",
      "dsand=1500 #density of sand in kg/m^3\n",
      "mstone=Vstone*dstone #volume of stone in m^3\n",
      "msand=Vsand*dsand #volume of sand in m^3\n",
      "Vtot=1 #total volume in m^3\n",
      "dair=1.1 #density of air in kg/m^3\n",
      "mair=Vair*dair #mass of air\n",
      "mtot=mair+msand+mliq+mstone #total mass in kg\n",
      "v=Vtot/mtot #specific volume in  m^3/kg\n",
      "d=1/v #overall density in kg/m^3\n",
      "print\"\\n hence,average specific volume is\",round(v,6),\"m^3/kg\" \n",
      "print\"\\n and overall density is\", round(d),\"kg/m^3\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " hence,average specific volume is 0.001325 m^3/kg\n",
        "\n",
        " and overall density is 755.0 kg/m^3\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2.3:pg-26"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#example 3\n",
      "#calculating the required force\n",
      "import math\n",
      "\n",
      "Dcyl=0.1 #cylinder diameter in m\n",
      "Drod=0.01 #rod diameter in m\n",
      "Acyl=math.pi*Dcyl**2/4 #cross sectional area of cylinder in m^2\n",
      "Arod=math.pi*Drod**2/4 #cross sectional area of rod in m^2\n",
      "Pcyl=250000 #inside hydaulic pressure in Pa\n",
      "Po=101000 #outside atmospheric pressure in kPa\n",
      "g=9.81 #acc. due to gravity in m/s^2\n",
      "mp=25 #mass of (rod+piston) in kg\n",
      "F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g #the force that rod can push within the upward direction in N\n",
      "print\"\\n hence,the force that rod can push within the upward direction is\",round(F,1),\" N\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " hence,the force that rod can push within the upward direction is 932.9  N\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2.4:pg-28"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#example 4\n",
      "#Calculating atmospheric pressure \n",
      "\n",
      "dm=13534 #density of mercury in kg/m^3\n",
      "H=0.750 #height difference between two columns in metres\n",
      "g=9.80665 #acc. due to gravity  in m/s^2\n",
      "Patm=dm*H*g/1000 #atmospheric pressure in kPa\n",
      "print\"\\n hence, atmospheric pressure is\",round(Patm,2),\"kPa\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " hence, atmospheric pressure is 99.54 kPa\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2.5:pg-28"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#example 5\n",
      "#pressure inside vessel\n",
      "\n",
      "dm=13590 #density of mercury in kg/m^3\n",
      "H=0.24 #height difference between two columns in metres\n",
      "g=9.80665 #acc. due to gravity  in m/s^2\n",
      "dP=dm*H*g #pressure difference in Pa\n",
      "Patm=13590*0.750*9.80665 #Atmospheric Pressure in Pa\n",
      "Pvessel=dP+Patm #Absolute Pressure inside vessel in Pa\n",
      "Pvessel=Pvessel/101325 #Absolute Pressure inside vessel in atm\n",
      "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,3),\"atm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " hence, the absolute pressure inside vessel is 1.302 atm\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2.6:pg-29"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#example 6\n",
      "#calculating pressure\n",
      "\n",
      "dg=750 #density of gaasoline in kg/m^3\n",
      "dR=1206 #density of R-134a in kg/m^3\n",
      "H=7.5 #height of storage tank in metres\n",
      "g=9.807 #acc. due to gravity in m/s^2\n",
      "dP1=dg*g*H/1000 #in kPa\n",
      "Ptop1=101 #atmospheric pressure in kPa\n",
      "P1=dP1+Ptop1\n",
      "print\"hence,pressure at the bottom of storage tank if fluid is gasoline is\",round(P1,1),\"kPa\" \n",
      "dP2=dR*g*H/1000 #in kPa\n",
      "Ptop2=1000 #top surface pressure in kPa\n",
      "P2=dP2+Ptop2\n",
      "print\"\\n hence, pressure at the bottom of storage tank if liquid is R-134a is\",round(P2),\"kPa\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "hence,pressure at the bottom of storage tank if fluid is gasoline is 156.2 kPa\n",
        "\n",
        " hence, pressure at the bottom of storage tank if liquid is R-134a is 1089.0 kPa\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2.7:pg-29"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#example 7\n",
      "#calculating balancing force\n",
      "\n",
      "Po=100#Outside atmospheric pressure in kPa\n",
      "F1=25 #net force on the smallest piston in kN\n",
      "A1=0.01 #cross sectional area of lower piston in m^2\n",
      "P1=Po+F1/A1 #fluid pressure in kPa\n",
      "d=900 #density of fluid in kg/m^3\n",
      "g=9.81 #acc. due to gravity in m/s^2\n",
      "H=6 #height of second piston in comparison to first one in m\n",
      "P2=P1-d*g*H/1000 #pressure at higher elevation on piston 2 in kPa\n",
      "A2=0.05 #cross sectional area of higher piston in m^3\n",
      "F2=(P2-Po)*A2 #balancing force on second piston in kN\n",
      "print\"\\n hence, balancing force on second larger piston is\",round(F2,1),\"N\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " hence, balancing force on second larger piston is 122.4 N\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}