path: root/Fundamentals_Of_Thermodynamics/Chapter6_6.ipynb

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  {

   "cells": [

    {

     "cell_type": "heading",

     "level": 1,

     "metadata": {},

     "source": [

      "Chapter6:FIRST-LAW ANALYSIS FOR A CONTROL VOLUME"

     ]

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.1:pg-182"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 1\n",

      "#calculating mass flow rate in kg/s\n",

      "import math\n",

      "R=0.287 #in kJ/kg-K\n",

      "T=25 #temperature in celsius\n",

      "P=150 #pressure in kPa\n",

      "v=R*(T+273.15)/P #specific volume in m^3/kg\n",

      "D=0.2 #diameter of pipe in metre\n",

      "A=math.pi*D**2/4 #cross sectional area in m^2\n",

      "V=0.1 #velocity of air in m/s\n",

      "m=V*A/v #mass flow rate in kg/s\n",

      "print\"\\n hence,the mass flow rate is\",round(m,4),\"kg/s\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence,the mass flow rate is 0.0055 kg/s\n"

       ]

      }

     ],

     "prompt_number": 11

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.2:pg-184"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 2\n",

      "#work done for adding the fluid\n",

      "\n",

      "P=600 #pressure in kPa\n",

      "m=1 #in kg\n",

      "v=0.001 #specific volume in m^3/kg\n",

      "W=P*m*v #necessary work in kJ for adding the fluid \n",

      "print\" \\n hence,the work involved in this process is\",round(W,4),\"kJ\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "  \n",

        " hence,the work involved in this process is 0.6 kJ\n"

       ]

      }

     ],

     "prompt_number": 13

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.3:pg-188"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 3\n",

      "#rate of flow of water\n",

      "\n",

      "hir=441.89 #in kJ/kg for refrigerant using steam table \n",

      "her=249.10 #in kJ/kg for refrigerant using steam table\n",

      "hiw=42 #in kJ/kg for water using steam table\n",

      "hew=83.95 #in kJ/kg for water using steam table\n",

      "mr=0.2 #the rate at which refrigerant enters the condenser in kg/s\n",

      "mw=mr*(hir-her)/(hew-hiw) #rate of flow of water in kg/s\n",

      "print\"\\n hence,the rate at which cooling water flows thorugh the condenser is\",round(mw,3),\"kg/s\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence,the rate at which cooling water flows thorugh the condenser is 0.919 kg/s\n"

       ]

      }

     ],

     "prompt_number": 15

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.4:pg-190"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 4\n",

      "#determining quality of steam\n",

      "\n",

      "hi=2850.1 #initial specific heat enthalpy for steam in kJ/kg\n",

      "Vi=50 #initial velocity of steam in m/s\n",

      "Ve=600 #final velocity of steam in m/s\n",

      "he=hi+Vi**2/(2*1000)-Ve**2/(2*1000) #final specific heat enthalpy for steam in kJ/kg\n",

      "hf=467.1 #at final state in kJ/kg\n",

      "hfg=2226.5 #at final state in kJ/kg\n",

      "xe=(he-hf)/hfg #quality of steam in final state\n",

      "print\" \\n hence, the quality is\",round(xe,2)"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " \n",

        " hence, the quality is 0.99\n"

       ]

      }

     ],

     "prompt_number": 19

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.4E:pg-191"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 4E\n",

      "#determining quality of steam\n",

      "\n",

      "hi=1227.5 #initial specific heat enthalpy for steam in Btu/lbm\n",

      "Vi=200 #initial velocity of steam in ft/s\n",

      "Ve=2000 #final velocity of steam in ft/s\n",

      "he=hi+Vi**2/(2*32.17*778)-Ve**2/(2*32.17*778) #final specific heat enthalpy for steam in Btu/lbm\n",

      "hf=198.31 #at final state in Btu/lbm\n",

      "hfg=958.81 #at final state in Btu/lbm\n",

      "xe=(he-hf)/hfg #quality of steam in final state\n",

      "print\" \\n hence, the quality is\",round(xe,2)"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " \n",

        " hence, the quality is 0.99\n"

       ]

      }

     ],

     "prompt_number": 2

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.5:pg-193"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 5\n",

      "#quality of ammonia leaving expansion valve\n",

      "\n",

      "hi=346.8 #specific heat enthalpy for ammonia at initial state in kJ/kg\n",

      "he=hi #specific heat enthalpy for ammonia at final state will be equal that at initial state because it is a throttling process\n",

      "hf=134.4 #at final state in kJ/kg\n",

      "hfg=1296.4#at final state in kJ/kg\n",

      "xe=(he-hf)*100/hfg #quality at final state\n",

      "print\"\\n hence,quality of the ammonia leaving the expansion valve is\",round(xe,2),\"%\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence,quality of the ammonia leaving the expansion valve is 16.38 %\n"

       ]

      }

     ],

     "prompt_number": 23

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.6:pg-194"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 6\n",

      "#power output of turbine in kW\n",

      "\n",

      "hi=3137 #initial specific heat of enthalpy in kJ/kg\n",

      "he=2675.5 #final specific heat of enthalpy in kJ/kg\n",

      "Vi=50.0 #initial velocity of steam in m/s\n",

      "Ve=100 #final velocity of steam in m/s\n",

      "Zi=6 #height of inlet conditions in metres\n",

      "Ze=3 #height of exit conditions in metres\n",

      "m=1.5 #mass flow rate of steam in kg/s\n",

      "g=9.8066 #acc. due to gravity in m/s^2\n",

      "Qcv=-8.5 #heat transfer rate from turbine in kW\n",

      "Wcv=Qcv+m*(hi+Vi**2/(2*1000)+g*Zi/1000)-m*(he+Ve**2/(2*1000)+g*Ze/1000) #power output of turbine in kW\n",

      "print\"\\n hence,the power output of the turbine is\",round(Wcv,1),\"kW\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence,the power output of the turbine is 678.2 kW\n"

       ]

      }

     ],

     "prompt_number": 28

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.7:pg-196"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 7\n",

      "#heat transfer rate in aftercooler\n",

      "\n",

      "V1=0 #we assume initial velocity to be zero because its given that it enters with a low velocity\n",

      "V2=25.0 #final velocity with which carbon dioxide exits in m/s\n",

      "h2=401.52 #final specific enthalpy of heat when carbon dioxide exits in kJ/kg\n",

      "h1=198 #initial specific enthalpy of heat in kJ/kg\n",

      "w=h1-h2-V2**2/(2*1000) #in kJ/kg\n",

      "Wc=-50 #power input to the compressor in kW\n",

      "m=Wc/w #mass flow rate of carbon dioxide in kg/s\n",

      "h3=257.9 #final specific enthalpy of heat when carbon dioxide flows into a constant pressure aftercooler\n",

      "Qcool=-m*(h3-h2) #heat transfer rate in the aftercooler in kW\n",

      "print\" \\n hence,heat transfer rate in the aftercooler is\",round(Qcool,1),\"kW\" \n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " \n",

        " hence,heat transfer rate in the aftercooler is 35.2 kW\n"

       ]

      }

     ],

     "prompt_number": 30

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.8:pg-197"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 8\n",

      "#Required pump work\n",

      "\n",

      "m=1.5 #mass flow rate of water in kg/s\n",

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

      "Zin=-15 #depth of water pump in well in metres\n",

      "Zex=0 #in metres\n",

      "v=0.001001 #specific volume in m^3/kg\n",

      "Pex=400+101.3 #exit pressure in kPa\n",

      "Pin=90 #in kPa\n",

      "W=m*(g*(Zin-Zex)*0.001-(Pex-Pin)*v) #power input in kW\n",

      "print\" \\n Hence, the pump requires power of input is\",-round(W,2),\"kW\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " \n",

        " Hence, the pump requires power of input is 0.84 kW\n"

       ]

      }

     ],

     "prompt_number": 39

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.9:pg-198"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 9\n",

      "#heat tranfer in simple steam power plant\n",

      "\n",

      "h1=3023.5 #specific heat of enthalpy of steam leaving boiler in kJ/kg\n",

      "h2=3002.5 #specific heat of enthalpy of steam entering turbine in kJ/kg\n",

      "x=0.9 #quality of steam entering condenser\n",

      "hf=226 #in kJ/kg\n",

      "hfg=2373.1 #in kJ/kg\n",

      "h3=hf+x*hfg #specific heat of enthalpy of steam entering condenser in kJ/kg\n",

      "h4=188.5 #specific heat of enthalpy of steam entering pump in kJ/kg\n",

      "q12=h2-h1 #heat transfer in line between boiler and turbine in kJ/kg\n",

      "w23=h2-h3 #turbine work in kJ/kg\n",

      "q34=h4-h3 #heat transfer in condenser\n",

      "w45=-4 #pump work in kJ/kg\n",

      "h5=h4-w45 #in kJ/kg\n",

      "q51=h1-h5 #heat transfer in boiler in kJ/kg\n",

      "print\"\\n hence, heat transfer in line between boiler and turbine is\",round(q12,1),\"kJ/kg\"  \n",

      "print\"\\n hence, turbine work is\",round(w23,1),\"kJ/kg\" \n",

      "print\"\\n hence, heat transfer in condenser is \",round(q34,1),\"kJ/kg\" \n",

      "print\"\\n hence, heat transfer in boiler is \",round(q51,1),\"kJ/kg\" "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence, heat transfer in line between boiler and turbine is -21.0 kJ/kg\n",

        "\n",

        " hence, turbine work is 640.7 kJ/kg\n",

        "\n",

        " hence, heat transfer in condenser is  -2173.3 kJ/kg\n",

        "\n",

        " hence, heat transfer in boiler is  2831.0 kJ/kg\n"

       ]

      }

     ],

     "prompt_number": 43

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.10:pg-200"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 10\n",

      "#analysis of refrigerator\n",

      "\n",

      "hf4=167.4 #in kJ/kg\n",

      "hfg4=215.6 #in kJ/kg\n",

      "h3=241.8 #specific heat of enthalpy of R-134a entering expansion valve\n",

      "h4=h3 #specific heat of enthalpy of R-134a leaving expansion valve\n",

      "h1=387.2 #in kJ/kg\n",

      "h2=435.1 #in kJ/kg\n",

      "x4=(h3-hf4)/hfg4 #quality of R-134a at evaporator inlet\n",

      "m=0.1 #mass flow rate in kg/s\n",

      "Qevap=m*(h1-h4) #rate of heat transfer to the evaporator\n",

      "Wcomp=-5 #power input to compressor in kW\n",

      "Qcomp=m*(h2-h1)+Wcomp #rate of heat transfer from compressor\n",

      "print\"\\n hence, the quality at the evaporator inlet is \",round(x4,3), \n",

      "print\"\\n hence, the rate of heat transfer to the evaporator is \",round(Qevap,2),\n",

      "print\"\\n hence, rate of heat transfer from the compressor is\",round(Qcomp,2), "

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " hence, the quality at the evaporator inlet is  0.345 \n",

        " hence, the rate of heat transfer to the evaporator is  14.54 \n",

        " hence, rate of heat transfer from the compressor is -0.21\n"

       ]

      }

     ],

     "prompt_number": 45

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.11:pg-204"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 11\n",

      "#Determining the final temperature of steam\n",

      "\n",

      "u2=3040.4 #final internal energy in kJ/kg\n",

      "hi=u2 #in kJ/kg\n",

      "P2=1.4 #final Pressure in MPa\n",

      "print\"Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \"\n",

      "T2=452 #final temperature in Celsius"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be \n"

       ]

      }

     ],

     "prompt_number": 46

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.12:pg-206"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 12\n",

      "#Calculating mass flow of steam in tank\n",

      "\n",

      "V1=0.4 #initial volume fo tank in m^3\n",

      "v1=0.5243 #initial specific volume in m^3/kg\n",

      "h1=3040.4 #initial specific enthalpy in kJ/kg\n",

      "u1=2548.9 #initial specific internal energy in kJ/kg\n",

      "m1=V1/v1 #initial mass of steam in tank in kg\n",

      "V2=0.4 #final volume in m^3\n",

      "print\"let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\" \n",

      "T2=342 #final temperature in Celsius\n",

      "v2=0.1974 #final specific volume in m^3/kg\n",

      "m2=V2/v2 #final mass of the steam in the tank in kg\n",

      "m=m2-m1 #mass of steam that flowsinto the tank\n",

      "print\" \\n Hence,mass of the steam that flows into the tank is\",round(m,3),\"kg\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation\n",

        " \n",

        " Hence,mass of the steam that flows into the tank is 1.263 kg\n"

       ]

      }

     ],

     "prompt_number": 48

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.13:pg-207"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 13\n",

      "#Calculating mass flow of steam in tank\n",

      "\n",

      "vf1=0.001725 #in m^3/kg\n",

      "vf2=0.0016 #in m^3/kg\n",

      "uf1=368.7 #in kJ/kg\n",

      "uf2=226 #in kJ/kg\n",

      "vg1=0.08313 #in m^3/kg\n",

      "vfg2=0.20381\n",

      "ug1=1341 #in kJ/kg\n",

      "ufg2=1099.7 #in kJ/kg\n",

      "Vf=1 #initial volume of liquid in m^3\n",

      "Vg=1 #initial volume of vapor in m^3\n",

      "mf1=Vf/vf1 #initial mass of liquid in kg\n",

      "mg1=Vg/vg1 #initial mass of vapor in kg\n",

      "m1=mf1+mg1 #initial mass of liquid in kg\n",

      "he=1461.1 #in kJ/kg\n",

      "V=2 #volume of tank in m^3\n",

      "print\"m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\"\n",

      "print\"Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\"\n",

      "x2=((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) #quality of ammonia\n",

      "v2=0.00160+(0.20381*x2) #final specific volume in m^3/kg\n",

      "m2=V/v2 #final mass of ammonia in kg\n",

      "m=m1-m2 #mass of ammonia withdrawn\n",

      "print\" \\n Hence,mass of ammonia withdrawn is\",round(m,1),\"kg\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.\n",

        "Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\n",

        " \n",

        " Hence,mass of ammonia withdrawn is 72.7 kg\n"

       ]

      }

     ],

     "prompt_number": 50

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex6.13E:pg-209"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#example 13E\n",

      "#Calculating mass flow of steam in tank\n",

      "\n",

      "vf1=0.02747#in ft3/lbm\n",

      "vf2=0.02564#in ft3/lbm\n",

      "uf1=153.89#in Btu/lbm\n",

      "uf2=97.16#in Btu/lbm\n",

      "vg1=1.4168 #in ft^3/lbm\n",

      "vfg2=3.2647#in ft^3/lbm\n",

      "ug1=576.23#in Btu/lbm\n",

      "ufg2=472.78#in Btu/lbm\n",

      "Vf=25 #initial volume of liquid in ft^3\n",

      "Vg=25 #initial volume of vapor in ft^3\n",

      "mf1=Vf/vf1 #initial mass of liquid in lbm\n",

      "mg1=Vg/vg1 #initial mass of vapor in lbm\n",

      "m1=mf1+mg1 #initial mass of liquid in lbm\n",

      "he=628 #in Btu/lbm\n",

      "V=50 #volume of tank in ft^3\n",

      "print\"m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.02564+3.2647*x2) and u2=uf2+x2*ufg2=97.16+3.2647*x2.\"\n",

      "print\"Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\"\n",

      "x2=((50*628)-(50*97.16)-(432391*0.02564))/((432391*3.2647)+(50*472.98)) #quality of ammonia\n",

      "x2=round(x2,6)\n",

      "v2=0.02564+(3.2647*x2) #final specific volume in ft^3/lbm\n",

      "v2=round(v2,6)\n",

      "m2=V/v2 #final mass of ammonia in lbm\n",

      "m=m1-m2 #mass of ammonia withdrawn\n",

      "print\" \\n Hence,mass of ammonia withdrawn is\",round(m,1),\"lbm\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.02564+3.2647*x2) and u2=uf2+x2*ufg2=97.16+3.2647*x2.\n",

        "Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.\n",

        " \n",

        " Hence,mass of ammonia withdrawn is 105.3 lbm\n"

       ]

      }

     ],

     "prompt_number": 27

    }

   ],

   "metadata": {}

  }

 ]

}