path: root/Thermodynamics_Demystified_by_Merle_C._Potter/Chapter4_2.ipynb

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  {

   "cells": [

    {

     "cell_type": "heading",

     "level": 1,

     "metadata": {},

     "source": [

      "Chapter 4:The First Law of Thermodynamics"

     ]

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.1:PG-62"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#initialization of variables\n",

      "K=100 # spring constant in kN/m\n",

      "d=0.8 # dispacement of spring in m\n",

      " # to get total work we integrate from 0 to 0.8 displacement\n",

      "x1=0; # lower limit of integration\n",

      "x2=0.8;  # upper limit of integration\n",

      "from scipy.integrate import quad\n",

      "\n",

      "# we find work\n",

      "def integrand(x,K):\n",

      "    return K*x\n",

      "\n",

      "W12, err = quad(integrand, x1, x2, K) # integrating to get work\n",

      "Q12=W12; # by first law of thermodynamics\n",

      "print \"The Heat transfer is \",int(Q12),\" J\"\n",

      "\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The Heat transfer is  32  J\n"

       ]

      }

     ],

     "prompt_number": 5

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.2:PG-65"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

      "P= 5*746 # power of fan converted in watt\n",

      "t=1*60*60 # time converted to seconds\n",

      "\n",

      "# by first law of thermodynamics Q=delU + W\n",

      "# Q=0 hence -W=delU\n",

      "# first we find work input\n",

      "W=-P*t # work in J\n",

      "delU=-W # from 1st law\n",

      "print \"The internal energy increase is \",float(delU),\" J\"\n",

      "# The answer is approximated in textbook\n",

      "# our answer is precise\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The internal energy increase is  13428000.0  J\n"

       ]

      }

     ],

     "prompt_number": 10

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.3:PG-65"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

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

      "T1=200 # initial temperature in degree celsius\n",

      "V1= 2 # initial volume in m^3\n",

      "Q=3500 # heat added in kJ\n",

      "v1=0.5342 # specific volume of steam at 200 degree celcius and 0.4 Mpa pressure from table C.3\n",

      "u1=2647 # specific internal energy in kJ/kg @ pressure = 0.4 MPa\n",

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

      "# we have a relation Between u2 and v2 from 1st law of thermodynamics\n",

      "v2=1.06 # specific volume at state 2 by trial and error and interpolation\n",

      "V2=m*v2 \n",

      "u2=((3500-400*(V2-V1))/m)+2647 # specific internal energy for v2=1.06 by trial and error\n",

      "\n",

      "# on interpolation from steam table at 0.4 MPa we get temperature \n",

      "T2=644 # temperature in degree celsius\n",

      "print \"The temperature for u2=\",round(u2),\" kJ and  v2 =\",round(v2,3),\" kg/m^3 is \\n \",int(T2),\" degree celsius\"\n",

      "# this numerical is solved by trial and error thus refer to Appendix C"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The temperature for u2= 3372.0  kJ and  v2 = 1.06  kg/m^3 is \n",

        "  644  degree celsius\n"

       ]

      }

     ],

     "prompt_number": 17

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.4:PG-67"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "\n",

      "# initialization of variables\n",

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

      "T1=200 # initial tmperature in degree celsius\n",

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

      "Q=3500 # heat added in kJ\n",

      "\n",

      "#solution\n",

      "h1=2860 # initial enthalpy @ 200*C and 400 kPa from steam table\n",

      "v=0.5342 # specific volume from steam table C.3 \n",

      "m=V/v;\n",

      "h2=(Q/m)+h1; # final enthalpy in kJ/kg from energy equation\n",

      "\n",

      "# NOW USING THIS ENTHLAPY AND INTERPOlATING FROM STEAM TABLE\n",

      "\n",

      "T2=600+(92.6/224)*100\n",

      "\n",

      "print \"The Final temperature is  \",int(T2),\" degree Celsius\"\n",

      "# result is obtained from interpolation on steam table\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The Final temperature is   641  degree Celsius\n"

       ]

      }

     ],

     "prompt_number": 20

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.5:PG-71"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

      "T1=300 # initial temperature in degree celsius\n",

      "T2=700 # final temperature in degree celsius\n",

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

      "m=3 # mass of steam in kg\n",

      "\n",

      "# solution\n",

      "# part (a)\n",

      "from scipy.integrate import quad\n",

      "\n",

      "# now we make function to integrate\n",

      "def integrand(T):\n",

      "    return 2.07+(T-400)/1480\n",

      "\n",

      "I, err = quad(integrand, T1, T2) # integrating specific heat over temperature range\n",

      "delH=m*I #integrate('2.07+(T-400)/1480','T',T1,T2) # expressing as function of temperature and integrating\n",

      "\n",

      "print\" The change in Enthalpy is \",int(delH),\" kJ \\n\"\n",

      " \n",

      "# part(b)\n",

      "CPavg=delH/(m*(T2-T1)) # avg value of specific heat at constant pressure\n",

      "print \" The average value of Cp is \",round(CPavg,2),\" kJ/kg.*C\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The change in Enthalpy is  2565  kJ \n",

        "\n",

        " The average value of Cp is  2.14  kJ/kg.*C\n"

       ]

      }

     ],

     "prompt_number": 24

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.6,PG-72"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "m=1 # mass of nitrogen in kg\n",

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

      "T2=1200 # final temperature in Kelvin\n",

      "M=28.0 # in kg/kmol\n",

      "# part(a)\n",

      "# the enthalpy change is found from gas table in App.E\n",

      "delh=36777-8723 # from gas table\n",

      "delH=delh/M \n",

      "print \" The entalpy change from gas table is \",round(delH),\" kJ/kg \\n\"\n",

      "\n",

      "# part (b) \n",

      "Cp=1.042 # from table B.2\n",

      "delH=Cp*(T2-T1)\n",

      "print \" The entalpy change  by assuming constant specific heat is \",round(delH),\" kJ/kg\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The entalpy change from gas table is  1002.0  kJ/kg \n",

        "\n",

        " The entalpy change  by assuming constant specific heat is  938.0  kJ/kg\n"

       ]

      }

     ],

     "prompt_number": 29

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.7:PG-76"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

      "x=0.7 # quality of steam\n",

      "P1=200 # initial pressure in kPa\n",

      "P2=800 # final pressure in kPa\n",

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

      "# The values are taken from TABLE C.2\n",

      "vf1=0.0010 # specific volume of saturated liquid at 200 kPa\n",

      "vg1=0.8857 # specific volume of saturated gas at 200 kPa\n",

      "uf1=504.5  # specific internal energy of saturated liquid @ state 1\n",

      "ug1=2529.5 # speciific internal energy of saturated gas @ state 1\n",

      "\n",

      "v1=vf1+x*(vg1-vf1); # specific volume of vapour\n",

      "m=V/v1\n",

      "\n",

      "u1=uf1+x*(ug1-uf1) # specific internal energy of vapour @ state 1\n",

      "v2=v1 # constant volume process\n",

      "u2=((0.6761-0.6203)*(3661-3853)/(0.6761-0.6181))+3853 # from steam table @ 800kPa by interpolating\n",

      "Q=m*(u2-u1) # heat transfer\n",

      "print \"The heat transfer is \",round(Q,3),\" kJ\"\n",

      "# The answer in the textbook is approximated"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The heat transfer is  5630.537  kJ\n"

       ]

      }

     ],

     "prompt_number": 34

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.8:PG-76"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

      "V=0.02 # volume in m^3\n",

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

      "T1=50+273 # initial temperature in kelvin\n",

      "T2=700+273 # final temperature in kelvin\n",

      "Q=50 # heat added in kJ\n",

      "R=287 # constant for air\n",

      "Cp=1 # constant for specific heat of air\n",

      "\n",

      "# using the ideal gas equation\n",

      "\n",

      "m=P*1000*V/(R*T1) # mass of air in kg\n",

      "W=Q-(m*Cp*(T2-T1)) # work done from first law\n",

      "# result\n",

      "print \"The Paddle work is \",round(W,2),\" kJ\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The Paddle work is  -6.09  kJ\n"

       ]

      }

     ],

     "prompt_number": 37

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.9,PG-77"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

      "V1=2 # initial volume in m^3\n",

      "V2=0.2 # final volume in m^3\n",

      "T1=20+273 # temperature in kelvin\n",

      "P=200 # pressure in kPa\n",

      "R=0.287 # constant for air\n",

      "gama=1.4 # polytropic index for air\n",

      "Cv=0.717 # specific heat at constant volume for air\n",

      "\n",

      "#solution\n",

      "\n",

      "#using the ideal gas equation\n",

      "m=(P*V1)/(R*T1) # mass in kg\n",

      "# process is adiabatic thus\n",

      "T2=T1*((V1/V2)**(gama-1)) # final temperature\n",

      "\n",

      "W=-m*Cv*(T2-T1) # work from first law\n",

      "print \"The Work is \",int(W),\" kJ\"\n",

      "# solution is approximated in textbook"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The Work is  -1510  kJ\n"

       ]

      }

     ],

     "prompt_number": 41

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.10:PG-79"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "# initialization of variables\n",

      "P1=2000.0 # initial pressure in kPa\n",

      "T1=600.0 # initial temperature in degree celsius\n",

      "p2=600.0 # final pressure in kPa\n",

      "T2=200.0 #   final temperature in degree celsius\n",

      "d1=0.06 # diameter of inlet pipe in metre\n",

      "d2=0.120 # diameter of outlet pipe in metre\n",

      "V1=20.0 # velocity at inlet in m/s\n",

      "\n",

      "# solution\n",

      "# from superheat table C.3 values are noted\n",

      "v1=0.1996 # specific volume of superheated steam @ 600*C and 2000 kPa\n",

      "v2=0.3520 # specific volume of superheated steam @ 200*C and 2000 kPa\n",

      "rho1=1/v1 #  initial density\n",

      "rho2=1/v2 # final density\n",

      "A1=(math.pi*d1**2)/4 # inlet area\n",

      "A2=(math.pi*d2**2)/4 # exit area\n",

      "\n",

      "V2=(rho1*A1*V1)/(rho2*A2) # from continuity equation\n",

      "print \" The Exit velocity is \",round(V2,2),\" m/s \\n\"\n",

      "\n",

      "mdot=rho1*A1*V1 # mass flow rate\n",

      "print\" The mass flow rate is \",round(mdot,3),\" kg/s\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The Exit velocity is  8.82  m/s \n",

        "\n",

        " The mass flow rate is  0.283  kg/s\n"

       ]

      }

     ],

     "prompt_number": 45

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.11:PG-82"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

      "P1=8000 # initial pressure in kPa\n",

      "T1=300  # temperature in degree celsius\n",

      "P2=2000 # final pressure in kPa\n",

      "\n",

      "# solution\n",

      "h1=2785 # specific enthalpy of steam in kJ/kg @ 8000 kPa and 300 degree celsius from  steam table\n",

      "h2=h1 # throttling process thus enthalpy is constant\n",

      "T2=212.4 # from steam table as we know enthalpy and pressure\n",

      "hf2=909 # specific enthalpy of saturated liquid @ 2000 kPa and 300 degree celsius\n",

      "hg2=2799.5 # specific enthalpy of saturated gas @ 2000 kPa and 300 degree celsius\n",

      "x2=(h2-hf2)/(hg2-hf2) # quality of steam\n",

      "\n",

      "vg2=0.0992 # specific    volume of saturated gas @ 2000 kPa and 212.4*c\n",

      "vf2=0.0012 # specific volume of saturated liquid @ 2000 kPa and 212.4*c\n",

      "v2=vf2+x2*(vg2-vf2) # from properties of pure substance\n",

      "\n",

      "print \"The Final Temperature and Specific volume is \",round(T2,1),\"*C and \",round(v2,3),\" m^3/kg\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The Final Temperature and Specific volume is  212.4 *C and  0.098  m^3/kg\n"

       ]

      }

     ],

     "prompt_number": 50

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.12:PG-84"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "# initialization of variables\n",

      "P1=4000 # inlet pressure in kPa\n",

      "T1=500 # inlet temperature in degree celsius\n",

      "V1=200 # inlet steam velocity in m/s\n",

      "d1=0.05 # inlet diameter in 'm'\n",

      "P2=80 # exit pressure in kPa\n",

      "d2=0.250 # exit diameter in 'm'\n",

      "\n",

      "# solution\n",

      "v1=0.08643 # specific volume from steam table @ 4000 kPa and 500*C\n",

      "v2=2.087 # specific volume from steam table @ 80 kPa and 500*C\n",

      "rho1=1/v1 # density at inlet\n",

      "rho2=1/v2 # density at outlet\n",

      "A1=(math.pi*d1**2)/4 # inlet area\n",

      "A2=(math.pi*d2**2)/4\n",

      "mdot=rho1*A1*V1 # mass flow rate\n",

      "mdot=round(mdot,3) # rounding to 3 significant digits\n",

      "\n",

      "#now using table C.3\n",

      "h1=3445 # initial specific enthalpy @ 4000 kPa and 500 *C \n",

      "h2=2666 # final specific enthalpy @ 80 kPa and 500 *C\n",

      "WT=-mdot*(h2-h1) # maximum power from first law\n",

      "print \" The power output is \",round(WT),\" kJ/s \\n \"\n",

      "\n",

      "V2=(A1*V1*rho1)/(A2*rho2) \n",

      "V2=round(V2) # rounding of digits\n",

      "delKE=mdot*((V2**2)-(V1**2))/2 # the change in kinetic energy\n",

      "print \" The change in K.E is \",round(delKE),\" J/s\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The power output is  3540.0  kJ/s \n",

        " \n",

        " The change in K.E is  -6250.0  J/s\n"

       ]

      }

     ],

     "prompt_number": 78

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.13:PG-85"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "# initialization of variables\n",

      "Wdot=10 # pump power in hp\n",

      "g=9.81 # acceleration due to gravity\n",

      "rho=1000 # density of water in kg/m^3\n",

      "d1=0.06 # inlet dimeter in 'm'\n",

      "d2=0.10 # oulet diamter in 'm'\n",

      "V1=10 # velocity of water at inlet in m/s\n",

      "\n",

      "#solution\n",

      "A1=math.pi*(d1**2)/4 # area of inlet\n",

      "A2=math.pi*(d2**2)/4 # area of outlet\n",

      "V2=A1*V1/A2 # oulet velocity from continuity equation\n",

      "\n",

      "mdot=rho*A1*V1 # mass flow rate\n",

      "delP=((((Wdot*746)/mdot)-((V2**2)-V1**2)/(2*g))*rho)/1000 # change in pressure in kPa\n",

      "print \"The rise in pressure is \",round(delP),\" kPa\"\n",

      "# The answer is approximated in textbook , our answer is precise \n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The rise in pressure is  268.0  kPa\n"

       ]

      }

     ],

     "prompt_number": 80

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.14:PG-85"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "\n",

      "# initialization of variables\n",

      "P1=7000.0 # inlet pressure in Pa\n",

      "T1=420.0 # inlet temperature in degree celsius\n",

      "V1=400.0 # inlet velocity in m/s\n",

      "d1=0.200 # inlet diameter in 'm'\n",

      "V2=700.0 # exit velocity in m/s\n",

      "k=1.4 # polytopic index for air\n",

      "Cp=1000 # specific heat at constant pressure for air in j/kg.K\n",

      "R=287 # specific gas constant for air\n",

      "\n",

      "#solution\n",

      "\n",

      "#part (a)\n",

      "T2=(((V1**2)-V2**2)/(2*Cp))+T1 # outlet temperature in degree celsius\n",

      "print \" The exit temperature is \",round(T2),\" *C \\n\"\n",

      "\n",

      "#part (b)\n",

      "\n",

      "rho1=P1/(R*(T1+273)) # density at entrance\n",

      "A1=(math.pi*d1**2)/4\n",

      "mdot=rho1*A1*V1 # \n",

      "print \" The mass flow rate is \",round(mdot,3),\" kg/s \\n\"\n",

      "\n",

      "# part (c)\n",

      "\n",

      "rho2=rho1*(((T2+273)/(T1+273))**(1/(k-1))) # density at exit\n",

      "# now we find the exit diameter\n",

      "d2=math.sqrt((rho1*V1*(d1)**2)/(rho2*V2))\n",

      "print \" The outlet diameter is \",round(d2,3),\" m\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The exit temperature is  255.0  *C \n",

        "\n",

        " The mass flow rate is  0.442  kg/s \n",

        "\n",

        " The outlet diameter is  0.212  m\n"

       ]

      }

     ],

     "prompt_number": 87

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex4.15:PG-89"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# initialization of variables\n",

      "mdots=100 # mass flow rate of sodium in kg/s\n",

      "Ts1=450 # inlet temperature of sodium in degree celsius\n",

      "Ts2=350 # exit temperature of sodium in degree celsius\n",

      "Cp=1.25 # specific heat of sodium in KJ/kg.*C\n",

      "Tw1=20 # inlet temperature of water in degree celsius\n",

      "Pw=5000 # inlet pressure of water in kPa \n",

      "\n",

      "# solution\n",

      "hw1=88.65 # enthalpy from table C.4\n",

      "hw2=2794 # enthalpy from table C.3\n",

      "mdotw=(mdots*Cp*(Ts1-Ts2))/(hw2-hw1) # mass flow rate of water\n",

      "print \" The mass flow rate of water is \",round(mdotw,2),\" kg/s \\n\"\n",

      "Qdot=mdotw*(hw2-hw1) # heat transfer in kW using energy equation\n",

      "# result\n",

      "print \" The rate of heat transfer is \",round(Qdot),\" kW\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The mass flow rate of water is  4.62  kg/s \n",

        "\n",

        " The rate of heat transfer is  12500.0  kW\n"

       ]

      }

     ],

     "prompt_number": 90

    }

   ],

   "metadata": {}

  }

 ]

}