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-{


- "metadata": {


-  "name": "",


-  "signature": "sha256:96f5bd0fa542764d6a8f75d5a1a42e4b79791697476f4029d0fa0fea1de65bed"


- },


- "nbformat": 3,


- "nbformat_minor": 0,


- "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": [


-      "Ex2.1E:pg-20"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#example 2\n",


-      "#weight of a person\n",


-      "\n",


-      "\n",


-      "m=1 #lbm\n",


-      "g=32 #acc.due to gravity in ft/s^2\n",


-      "F=m*g/32.174 #weight of 1 lbm mass in lbf\n",


-      "print\"\\n hence,weight of person is\",round(F,4),\" lbf\""


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        "\n",


-        " hence,weight of person is 0.9946  lbf\n"


-       ]


-      }


-     ],


-     "prompt_number": 3


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex2.2:pg-24"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#example 3\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 4\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 5\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 6\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": [


-      "Ex2.5E:pg-28"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#example 7\n",


-      "#pressure inside vessel\n",


-      "\n",


-      "dm=848 #density of mercury in lbm/ft^3\n",


-      "H=9.5  #height difference between two columns in inches\n",


-      "g=32.174 #acc. due to gravity  in ft/s^2\n",


-      "dP=dm*H*g #pressure difference in lbf/in^2\n",


-      "Patm=848*9.5*32.174/(1728*32.174) #Atmospheric Pressure in lbf/in^2\n",


-      "Pvessel=dP+Patm #Absolute Pressure inside vessel in lbf/in^2\n",


-      "Pvessel=848*29.5*32.174/(1728*32.174)+Patm #Absolute Pressure inside vessel in lbf/in^2\n",


-      "print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,2),\"lbf/in^2\""


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        "\n",


-        " hence, the absolute pressure inside vessel is 19.14 lbf/in^2\n"


-       ]


-      }


-     ],


-     "prompt_number": 9


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex2.6:pg-29"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#example 8\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 9\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


-    }


-   ],


-   "metadata": {}


-  }


- ]


-}
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