path: root/Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb

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 "worksheets": [

  {

   "cells": [

    {

     "cell_type": "heading",

     "level": 1,

     "metadata": {},

     "source": [

      "Chapter5:THE FIRST LAW OF THERMODYNAMICS"

     ]

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.1:pg-131"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 1\n",

      "#calculating height\n",

      "\n",

      "m=1100 #mass of car in kg\n",

      "ke=400 #kinetic energy of car in kJ\n",

      "V=(2*ke*1000/m)**0.5 #velocity of car in m/s\n",

      "g=9.807 #acc. due to gravity in m/s^2\n",

      "H=ke*1000/(m*g) #height to which the car  should be lifted so that its potential energy equals its kinetic energy\n",

      "print\"hence,the car should be raised to a height is\",round(H,1),\"m\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "hence,the car should be raised to a height is 37.1 m\n"

       ]

      }

     ],

     "prompt_number": 3

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.1E:pg-132"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 1E\n",

      "#calculating height\n",

      "\n",

      "m=2400 #mass of car in lbm\n",

      "ke=400 #kinetic energy of car in Btu\n",

      "V=(2*ke*778.17*32.174/m)**0.5 #velocity of car in ft/s\n",

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

      "H=ke*778.17*32.174/(m*g) #height to which the car  should be lifted so that its potential energy equals its kinetic energy\n",

      "print\"hence,the car should be raised to a height is\",round(H,1),\"ft\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "hence,the car should be raised to a height is 129.7 ft\n"

       ]

      }

     ],

     "prompt_number": 2

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.2:pg-134"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 2\n",

      "#change in internal energy\n",

      "\n",

      "W=-5090 #work input to paddle wheel in kJ\n",

      "Q=-1500 #heat transfer from tank in kJ\n",

      "dU=Q-W #change in internal energy in kJ\n",

      "print\"hence,change in internal energy is\",round(dU),\"kj\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "hence,change in internal energy is 3590.0 kj\n"

       ]

      }

     ],

     "prompt_number": 5

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.3:pg-134"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 3\n",

      "#analysis of energy transfer\n",

      "\n",

      "g=9.806 #acceleration due to gravity in m/s^2\n",

      "m=10 #mass of stone in kg\n",

      "H1=10.2 #initial height of stone above water in metres\n",

      "H2=0 #final height in metres\n",

      "dKE1=-m*g*(H2-H1) #change in kinetic energy when stone enters state 2 in J\n",

      "dPE1=-1 #change in potential energy when stone enters state 2 in J\n",

      "print\"\\n hence,when stone is \",round(dKE1),\"J\"\n",

      "print\"\\n and change in potential energy is \",round(dPE1),\"J\"\n",

      "dPE2=0 #change in potential energy when stone enters state 3 in JQ2=0 //no heat transfer when stone enters state 3 in J\n",

      "W2=0 #no work done when stone enters state 3 in J\n",

      "dKE2=-1 #change in kinetic energy when stone enters state 3\n",

      "dU2=-dKE2 #change in internal energy when stone enters state 3 in J\n",

      "print\"\\n hence,when stone has just come to rest in the bucket is \",round(dKE2),\"J\" \n",

      "print\"\\n and dU is\",round(dU2),\"J\"  \n",

      "dKE3=0 #change in kinetic energy when stone enters state 4\n",

      "dPE=0 #change in potential energy when stone enters state 4 in J\n",

      "W3=0 #no work done when stone enters state 4 in J\n",

      "dU3=-1 #change in internal energy when stone enters state 4 in J\n",

      "Q3=dU3 #heat transfer when stone enters state 4 in J\n",

      "print\"\\n hence,when stone has entered state 4 is\",round(dU3),\"J\" \n",

      "print\"\\n and Q3 is \",round(Q3),\"J\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " \n",

        " hence,when stone is  1000.0 J\n",

        "\n",

        " and change in potential energy is  -1.0 J\n",

        "\n",

        " hence,when stone has just come to rest in the bucket is  -1.0 J\n",

        "\n",

        " and dU is 1.0 J\n",

        "\n",

        " hence,when stone has entered state 4 is -1.0 J\n",

        "\n",

        " and Q3 is  -1.0 J\n"

       ]

      }

     ],

     "prompt_number": 7

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.4:pg-136"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 4\n",

      "#Determinig the missing properties \n",

      "\n",

      "T1=300 #given temp. in Celsius\n",

      "u1=2780 #given specific internal enrgy in kJ/kg\n",

      "print\"From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\" #hiven pressure in kPa\n",

      "u2=2000 #given specific intrernal energy in kJ/kg\n",

      "print\"at P=2000 kPa\"\n",

      "uf=906.4 #in kJ/kg\n",

      "ug=2600.3 #in kJ/kg \n",

      "x2=(u2-906.4)/(ug-uf) \n",

      "print\"Also, under the given conditions\"\n",

      "vf=0.001177 #in m^3/kg \n",

      "vg=0.099627 #in m^3/kg\n",

      "v2=vf+x2*(vg-vf)#Specific volume for water in m^3/kg\n",

      "print\"\\n hence,specific volume for water is \",round(v2,5),\"m^3/kg\" \n",

      "print\"\\n Therefore ,this state is \",round(x2,4),\"N\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\n",

        "at P=2000 kPa\n",

        "Also, under the given conditions\n",

        "\n",

        " hence,specific volume for water is  0.06474 m^3/kg\n",

        "\n",

        " Therefore ,this state is  0.6456 N\n"

       ]

      }

     ],

     "prompt_number": 12

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.5:pg-138"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 5\n",

      "#calculating heat transfer for the given process\n",

      "\n",

      "Vliq=0.05 #volume of saturated liquid in m^3\n",

      "vf=0.001043 #in m^3/kg\n",

      "Vvap=4.95 #volume of saturated water vapour in m^3\n",

      "vg=1.6940 #in m^3/kg\n",

      "m1liq=Vliq/vf #mass of liquid in kg\n",

      "m1liq=round(m1liq,2)\n",

      "m1vap=Vvap/vg #mass of vapors in kg\n",

      "m1vap=round(m1vap,2)\n",

      "u1liq=417.36 #specific internal energy of liquid in kJ/kg\n",

      "u1vap=2506.1 #specific internal energy of vapors in kJ/kg\n",

      "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in kJ\n",

      "m=m1liq+m1vap #total mass in kg\n",

      "V=5.0 #total volume in m^3\n",

      "v2=V/m #final specific volume in m^3/kg\n",

      "print\"by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\"\n",

      "u2=2600.5 #specific internal energy at final state in kJ/kg\n",

      "U2=m*u2 #internal energy at final state in kJ\n",

      "Q=U2-U1 #heat transfer for the process in kJ\n",

      "print\"\\n hence,heat transfer for the process is\",round(Q),\"kJ\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\n",

        "\n",

        " hence,heat transfer for the process is 104935.0 kJ\n"

       ]

      }

     ],

     "prompt_number": 36

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.5E:pg-140"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 5E\n",

      "#calculating heat transfer for the given process\n",

      "\n",

      "Vliq=1 #volume of saturated liquid in ft^3\n",

      "vf=0.01672 #in ft^3/lbm\n",

      "Vvap=99 #volume of saturated water vapour in m^3\n",

      "vg=26.80 #in ft^3/lbm\n",

      "m1liq=Vliq/vf #mass of liquid in lbm\n",

      "m1liq=round(m1liq,2)\n",

      "m1vap=Vvap/vg #mass of vapors in lbm\n",

      "m1vap=round(m1vap,2)\n",

      "u1liq=180.1 #specific internal energy of liquid in Btu/lbm\n",

      "u1vap=1077.6 #specific internal energy of vapors in Btu/lbm\n",

      "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in Btu\n",

      "U1=round(U1)\n",

      "m=m1liq+m1vap #total mass in lbm\n",

      "V=100.0 #total volume in ft^3\n",

      "v2=V/m #final specific volume in ft^3/lbm\n",

      "print\"by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\"\n",

      "u2=1117.0 #specific internal energy at final state in Btu/lbm\n",

      "U2=m*u2 #internal energy at final state in Btu\n",

      "U2=round(U2)\n",

      "Q=U2-U1 #heat transfer for the process in Btu\n",

      "print\"\\n hence,heat transfer for the process is\",round(Q),\"Btu\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\n",

        "\n",

        " hence,heat transfer for the process is 56182.0 Btu\n"

       ]

      }

     ],

     "prompt_number": 23

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.6:pg-143"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 6\n",

      "#calculating work and heat transfer for the process\n",

      "\n",

      "V1=0.1 #volume of cylinder in m^3\n",

      "m=0.5 #mass of steam in kg\n",

      "v1=V1/m #specific volume of steam in m^3/kg\n",

      "vf=0.001084 #m^3/kg\n",

      "vfg=0.4614 #m^3/kg\n",

      "x1=(v1-vf)/vfg #quality\n",

      "hf=604.74 #kJ/kg\n",

      "hfg=2133.8#kJ/kg\n",

      "h2=3066.8 #final specific heat enthalpy in kJ/kg\n",

      "h1=hf+x1*hfg #initial specific enthalpy in kJ/kg\n",

      "Q=m*(h2-h1) #heat transfer for this process in kJ\n",

      "P=400 #pressure inside cylinder in kPa\n",

      "v2=0.6548 #specific enthalpy in m^3/kg\n",

      "W=m*P*(v2-v1) #work done for the process in kJ\n",

      "print\"\\n hence, work done for the process is\",round(W),\"kJ\" \n",

      "print\"\\n and heat transfer is \",round(Q,1),\"kJ\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence, work done for the process is 91.0 kJ\n",

        "\n",

        " and heat transfer is  771.1 kJ\n"

       ]

      }

     ],

     "prompt_number": 1

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.7:pg-144"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# example 7\n",

      "#calculating work and heat transfer for the process\n",

      "V1=0.010 #volume of cylinder in m^3\n",

      "V2=0.040 # m^3\n",

      "P1=573 #kPa\n",

      "v1=0.03606 #specific volume of steam in m^3/kg\n",

      "u1=389.2 #kJ/kg\n",

      "m=V1/v1#mass of steam in kg\n",

      "v2=v1*(V2/V1)\n",

      "P2=163 #kPa\n",

      "u2=395.8 # kJ/kg\n",

      "W=8 # kJ\n",

      "Q2=m*(u2-u1)+W\n",

      "u3=383.4#kJ/kg\n",

      "W2=m*(u1-u3)#kJ\n",

      "print\"\\n and heat transfer is \",round(Q2,2),\"kJ\" \n",

      "print\"\\n hence, work done for the process is\",round(W2,1),\"kJ\" \n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " and heat transfer is  9.83 kJ\n",

        "\n",

        " hence, work done for the process is 1.6 kJ\n"

       ]

      }

     ],

     "prompt_number": 18

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.8:pg-151"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 8\n",

      "#calculating change in enthalpy\n",

      "import math\n",

      "\n",

      "h1=273.2 #specific heat enthalpy for oxygen at 300 K\n",

      "h2=1540.2 #specific heat enthalpy for oxygen at 1500 K\n",

      "T1=300 #initial temperature in K\n",

      "T2=1500 #final temparature in K\n",

      "\n",

      "dh1=h2-h1 #this change in specific heat enthalpy is calculated using ideal gas tables \n",

      "dh3=0.922*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K\n",

      "dh4=1.0767*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature\n",

      "print\"\\n Hence,change in specific heat enthalpy if ideal gas tables are used is \",round(dh1,1),\"kJ/kg\"\n",

      "print\"\\n if specific heat is assumed to be constant and using its value at T1 is\",round(dh3,1),\"kJ/kg\"\n",

      "print\"\\n if specific heat is assumed to be constant at its value at (T1+T2)/2 is\",round(dh4,1),\"kJ/kg\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " Hence,change in specific heat enthalpy if ideal gas tables are used is  1267.0 kJ/kg\n",

        "\n",

        " if specific heat is assumed to be constant and using its value at T1 is 1106.4 kJ/kg\n",

        "\n",

        " if specific heat is assumed to be constant at its value at (T1+T2)/2 is 1292.0 kJ/kg\n"

       ]

      }

     ],

     "prompt_number": 4

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.9:pg-152"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 9\n",

      "#determining amount of heat transfer\n",

      "\n",

      "P=150 #pressure of nitrogen in cylinder in kPa\n",

      "V=0.1 #initial volume of cylinder in m^3\n",

      "T1=25 #initial temperature of nitrogen in celsius\n",

      "T2=150 #final tempareture of nitrogen in celsius\n",

      "R=0.2968 #in kJ/kg-K\n",

      "m=P*V/(R*(T1+273)) #mass of nitrogen in kg\n",

      "Cv=0.745 #constant volume specific heat for nitrogen in kJ/kg-K\n",

      "W=-20 #work done on nitrogen gas in kJ\n",

      "Q=m*Cv*(T2-T1)+W #heat transfer during the process in kJ\n",

      "print\"\\n hence,the heat transfer for the above process is\",round(Q,1),\"kJ\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence,the heat transfer for the above process is -4.2 kJ\n"

       ]

      }

     ],

     "prompt_number": 43

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.9E:pg-153"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 9E\n",

      "#determining amount of heat transfer\n",

      "\n",

      "P=20 #pressure of nitrogen in cylinder in lbf/in^2\n",

      "V=2 #initial volume of cylinder in ft^3\n",

      "T1=80 #initial temperature of nitrogen in F\n",

      "T2=300 #final tempareture of nitrogen in F\n",

      "R=55.15 #in ft*lbf/lbm-R\n",

      "m=P*V*144/(R*(540)) #mass of nitrogen in lbm\n",

      "Cv=0.177 #constant volume specific heat for nitrogen in Btu/lbm-R\n",

      "W=-9.15 #work done on nitrogen gas in Btu\n",

      "Q=m*Cv*(T2-T1)+W #heat transfer during the process in Btu\n",

      "print\"\\n hence,the heat transfer for the above process is\",round(Q,2),\"Btu\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence,the heat transfer for the above process is -1.62 Btu\n"

       ]

      }

     ],

     "prompt_number": 31

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.10:pg-155"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 10\n",

      "#calculating rate of increase of internal energy\n",

      "\n",

      "W=-12.8*20 #power consumed in  J/s\n",

      "Q=-10 #heat transfer rate from battery in J/s\n",

      "r=Q-W #rate of increase of internal energy\n",

      "print\"\\n hence,the rate of increase of internal energy is\",round(r),\"J/s\"  "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence,the rate of increase of internal energy is 246.0 J/s\n"

       ]

      }

     ],

     "prompt_number": 45

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex5.11:pg-155"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 11\n",

      "#rate of change of temperature\n",

      "\n",

      "Q=1500.0 #power produced by burning wood in J/s\n",

      "mair=1 #mass of air in kg\n",

      "mwood=5 #mass of soft pine wood in kg \n",

      "miron=25 #mass of cast iron in kg\n",

      "Cvair=0.717 #constant volume specific heat for air in kJ/kg\n",

      "Cwood=1.38 #constant volume specific heat for wood in kJ/kg\n",

      "Ciron=0.42 #constant volume specific heat for iron in kJ/kg\n",

      "dT=75-20 #increase in temperature in Celsius\n",

      "T=(Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) #rate of change of temperature in K/s\n",

      "dt=(dT/T)/60 #in minutes\n",

      "print\" hence,the rate of change of temperature is\",round(T,4),\"k/s\"  \n",

      "print\" and time taken to reach a temperature of T is\",round(dt),\"min\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " hence,the rate of change of temperature is 0.0828 k/s\n",

        " and time taken to reach a temperature of T is 11.0 min\n"

       ]

      }

     ],

     "prompt_number": 49

    }

   ],

   "metadata": {}

  }

 ]

}