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   "source": [
    "# Chapter 12:Principles of mass transfer"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex12.1:pg-496"
   ]
  },
  {
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   "execution_count": 1,
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     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 12, Example 1\n",
      "The flow area is given by A=(pi*di**2)/4 in m**2\n",
      "3e-05\n",
      "The molar concentration of mixture which is constant throughout is given by c=p/(R*T)\n",
      "0.04079\n",
      "Nhe=Nair=(A*c*Db*(Yao-yal))/L in kmol/sec\n",
      "mass flow rate of helium is given by m=Mhe*Nhe in kg/sec \n",
      "1.7e-11\n",
      "mass flow rate of air is given by m=Mair*Nair in kg/sec \n",
      "1.2e-10\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"Introduction to heat transfer by S.K.Som, Chapter 12, Example 1\"\n",
    "#The pressure in the pipeline that transports helium gas at a rate of 4kg/s is maintained at pressure(p)=1 atm or 101*10**3 pascal.\n",
    "#The internal daimeter of tube is (di)=6mm or .006m\n",
    "#The temprature of both air and helium is (T)=25°C or 298 K.\n",
    "#The diffusion coefficient of helium in air at normal atmosphere is(Dab)=7.20*10**-5 m**2/s\n",
    "#The venting tube extends to a length(L)=20m in the atmosphere.\n",
    "di=.006;\n",
    "print \"The flow area is given by A=(pi*di**2)/4 in m**2\"\n",
    "A=(math.pi*di**2)/4\n",
    "print round(A,5)\n",
    "p=101*10**3;\n",
    "R=8.31*10**3;#gas constant\n",
    "T=298;\n",
    "Dab=7.20*10**-5;\n",
    "L=20;\n",
    "#c is the molar concentration\n",
    "print \"The molar concentration of mixture which is constant throughout is given by c=p/(R*T)\"\n",
    "c=p/(R*T)\n",
    "print round(c,5)\n",
    "#helium has been considered as species A so (helium mole fraction at the bottom of the tube)is Yao=1 and (helium mole fraction at the bottom of the tube)is Yal=0\n",
    "Yal=0;\n",
    "Yao=1;\n",
    "#Nhe and Nair are molar rate of helium and air respectively\n",
    "print \"Nhe=Nair=(A*c*Db*(Yao-yal))/L in kmol/sec\"\n",
    "Nair=(A*c*Dab*(Yao-Yal))/L\n",
    "Nhe=Nair;\n",
    "#Molecular weights of air and helium are 29kg/kmol and 4 kg/kmol respectively.\n",
    "Mhe=4;\n",
    "Mair=29;\n",
    "#mass flow rate of helium is mhe\n",
    "print \"mass flow rate of helium is given by m=Mhe*Nhe in kg/sec \"\n",
    "mhe=Mhe*Nhe\n",
    "print round(mhe,12)\n",
    "#mass flow rate of air is mair\n",
    "print \"mass flow rate of air is given by m=Mair*Nair in kg/sec \"\n",
    "mair=Mair*Nair\n",
    "print round(mair,11)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex12.2:pg-500"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
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     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 12, Example 2\n",
      "The film temperature is given by Tf=(T+Tw)/2 in °C \n",
      "30.0\n",
      "The density of water at bulb surface is given by rhos=(Ps*M)/(R*Ts) in kg/m**3 \n",
      "0.0173\n",
      "The concentration of water vapour at free stream is rhoinf=rhos-(hheat/hmass)*((Tinf-Ts)/hfg) in kg/m**3 \n",
      "0.00784\n",
      "The relative humidity is given by rehu=(rhoinf/rhosteam)*100 in percentage \n",
      "15.38028\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "print \"Introduction to heat transfer by S.K.Som, Chapter 12, Example 2\"\n",
    "#The temprature of atmospheric air (T)=40°C which flows over a wet bulb thermometer.\n",
    "#The reading of wet bulb thermometer which is called the wet bulb temprature is (Tw)=20°C\n",
    "T=40;\n",
    "Tw=20.0;\n",
    "#Tf is the film temprature\n",
    "print \"The film temperature is given by Tf=(T+Tw)/2 in °C \"\n",
    "Tf=(T+Tw)/2\n",
    "print round(Tf,5)\n",
    "Tinf=T;#surrounding temprature\n",
    "#The properties of air at film temprature are density(rho=1.13kg/m**3),specific heat(cp=1.007kJ/(kg*K)),Thermal diffusivity(alpha=0.241*10**-4m**2/s)\n",
    "#The diffusivity Dab=0.26*10**-4 m**2/s\n",
    "#The enthalpy of vaporisation of water at 20°C is hfg=2407kJ/kg or 2407*10**3 J/kg\n",
    "#The partial pressure of water vapour is the saturation pressure corresponding to 20°C so from steam table Ps=2.34kPa or 2.34*10**3 Pa.\n",
    "rho=1.13;\n",
    "cp=1.007*10**3;\n",
    "alpha=0.241*10**-4;\n",
    "Dab=0.26*10**-4;\n",
    "hfg=2407*10**3;\n",
    "Ps=2.34*10**3;\n",
    "#The temprature at bulb surface Ts=20°C or 293K\n",
    "Ts=Tw+273;#in kelvin\n",
    "R=8.31*10**3;#gas constant\n",
    "#The molecular weight of water is M=18\n",
    "M=18;\n",
    "#The density of water at bulb surface is rhos\n",
    "print \"The density of water at bulb surface is given by rhos=(Ps*M)/(R*Ts) in kg/m**3 \"\n",
    "rhos=(Ps*M)/(R*Ts)\n",
    "print round(rhos,5)\n",
    "#Let X=hheat/hmass=rho*cp*(alpha/Dab)**(2/3).\n",
    "X=rho*cp*(alpha/Dab)**(2/3);\n",
    "#At steady atate (Rate of heat transfer from air to wet cover of thermometer bulb)=(Heat removed by evaporation of water from the wet cover of thermometer bulb)\n",
    "#hheat*(Tinf-Ts)=hmass*(rhos-rhoinf)*hfg\n",
    "#Rearranging above we get rhoinf=rhos-(hheat/hmass)*((Tinf-Ts)/hfg)\n",
    "#The concentration of water vapour at free stream is rhoinf\n",
    "print \"The concentration of water vapour at free stream is rhoinf=rhos-(hheat/hmass)*((Tinf-Ts)/hfg) in kg/m**3 \"\n",
    "rhoinf=rhos-((X)*((Tinf-Tw)/hfg))\n",
    "print round(rhoinf,5)\n",
    "#The mass concentration of saturated water vapour(rhosteam) at 40°C(as found from steam table) is .051 kg/m**3\n",
    "rhosteam=.051;\n",
    "#The relative humidity is (rehu)\n",
    "print \"The relative humidity is given by rehu=(rhoinf/rhosteam)*100 in percentage \"\n",
    "rehu=(rhoinf/rhosteam)*100\n",
    "print round(rehu,5)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex12.3:pg-503"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
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   "outputs": [
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     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 12, Example 3\n",
      "The mole fraction of water vapour at the interface is given by Yao=pvapour/p\n",
      "0.03963\n",
      "The total molecular concentration (c) through the tube remains constant is given by c=p/(R*T) in kmol/m**3\n",
      "0.03286\n",
      "The cross sectional area of the tube is given by A=(pi*(di*10**-3)**2)/4 in m**2\n",
      "0.00096\n",
      "The molar flow rate of water vapour is given by N=mdot/M in kmol/s\n",
      "1e-10\n",
      "The diffusion coefficient of water vapour is Dab=(N*L)/(c*A*ln[(1-Yal)/(1-Yao)]) in m/s\n",
      "3e-05\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"Introduction to heat transfer by S.K.Som, Chapter 12, Example 3\"\n",
    "#The diameter of tube is (di)=35mm which measures binary diffusion coefficient of water vapour in air at temprature,T=20°C or 293 K.\n",
    "#The measurement is done at height of 1500 m where the atmospheric pressure is (p)=80kPa.\n",
    "p=80;\n",
    "T=293.0;\n",
    "#The distance from the water surface to the open end of the tube is L=500 mm or 0.5m.\n",
    "L=.5;\n",
    "#After t=12 days of continuous operation at constant pressure and temprature the amount of water evaporated was measured to be m= 1.2*10**-3kg.\n",
    "m= 1.2*10**-3;\n",
    "#From the steam table pvapour=3.17kPa\n",
    "pvapour=3.17;#partial pressure of vapour\n",
    "#Yao is the mole fraction of water vapour at the interface\n",
    "print \"The mole fraction of water vapour at the interface is given by Yao=pvapour/p\"\n",
    "Yao=pvapour/p\n",
    "print round(Yao,5)\n",
    "#The mole fraction of water vapour at the top end of the tube is Yal=0\n",
    "Yal=0;\n",
    "R=8.31*10**3;#gas constant\n",
    "#The total molecular concentration is (c)\n",
    "print \"The total molecular concentration (c) through the tube remains constant is given by c=p/(R*T) in kmol/m**3\"\n",
    "c=(p*10**3)/(R*T)\n",
    "print round(c,5)\n",
    "di=35;\n",
    "#A is the cross sectional area of the tube\n",
    "print \"The cross sectional area of the tube is given by A=(pi*(di*10**-3)**2)/4 in m**2\"\n",
    "A=(math.pi*(di*10**-3)**2)/4\n",
    "print round(A,5)\n",
    "#The molecular weight of wate is M=18\n",
    "M=18;\n",
    "#The mass flow rate is given by mdot=(m/(12*24*3600))\n",
    "mdot=(m/(12*24*3600));\n",
    "#N is the molar flow rate of water vapour\n",
    "print \"The molar flow rate of water vapour is given by N=mdot/M in kmol/s\"\n",
    "N=mdot/M\n",
    "print round(N,10)\n",
    "#The molar flow rate of water vapour can also be written as N=(c*Dab*A*ln[(1-Yal)/(1-Yao)])/L\n",
    "#The diffusion coefficient of water vapour is Dab=(N*L)/(c*A*ln[(1-Yal)/(1-Yao)])\n",
    "#let us take X=math.log10((1-Yal)/(1-Yao)) and Y=math.log10(2.7182)\n",
    "X=math.log10((1-Yal)/(1-Yao));\n",
    "Y=math.log10(2.7182);\n",
    "#ln[(1-Yal)/(1-Yao)] is given by\n",
    "ln=X/Y;\n",
    "print \"The diffusion coefficient of water vapour is Dab=(N*L)/(c*A*ln[(1-Yal)/(1-Yao)]) in m/s\"\n",
    "Dab=(N*L)/(c*A*ln)\n",
    "print round(Dab,5)"
   ]
  }
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