path: root/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter9.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 09:Properties of pure substances"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.1:pg-302"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.1\n",
      "\n",
      " At 1 MPa, \n",
      " saturation temperature is  179.91  degree celcius\n",
      "\n",
      " Changes in specific volume is  0.193313  m**3/kg\n",
      "\n",
      " Change in entropy during evaporation is  4.4478  kJ/kg K\n",
      "\n",
      " The latent heat of vaporization is  2015.3  kJ/kg\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# At 1 MPa\n",
    "tsat = 179.91 # Saturation temperature in degree Celsius\n",
    "vf = 0.001127 # Specific volume of fluid in m**3/kg\n",
    "vg = 0.19444 # Specific volume of gas in m**3/kg \n",
    "sf = 2.1387 # Specific entropy of fluid in kJ/kgK\n",
    "sg = 6.5865# Specific entropy of gas in kJ/kgK\n",
    "print \"\\n Example 9.1\"\n",
    "vfg = vg-vf # Change in specific volume due to evaporation\n",
    "sfg = sg-sf# Change in specific entropy due to evaporation\n",
    "hfg = 2015.3\n",
    "print \"\\n At 1 MPa, \\n saturation temperature is \",tsat ,\" degree celcius\"\n",
    "print \"\\n Changes in specific volume is \",vfg ,\" m**3/kg\"\n",
    "print \"\\n Change in entropy during evaporation is \",sfg ,\" kJ/kg K\"\n",
    "print \"\\n The latent heat of vaporization is \",hfg ,\" kJ/kg\"\n",
    "# Data is given in the table A.1(b) in Appendix in the book\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.2:pg-302"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.2\n",
      "\n",
      " pressure =  0.6  Mpa\n",
      " Temperature =  158.85  degree centigrade\n",
      " Specific volume =  0.3156  m**3/kg\n",
      " enthalpy =  2756.8  kJ/kg\n"
     ]
    }
   ],
   "source": [
    "# Given that\n",
    "s = 6.76 # Entropy of saturated steam in kJ/kgK\n",
    "print \"\\n Example 9.2\"\n",
    "# From the table A.1(b) given in the book at s= 6.76 kJ/kgK\n",
    "p = 0.6\n",
    "t=158.85\n",
    "v_g=0.3156\n",
    "h_g=2756.8\n",
    "print \"\\n pressure = \",p ,\" Mpa\\n Temperature = \",t ,\" degree centigrade\\n Specific volume = \",v_g ,\" m**3/kg\\n enthalpy = \",h_g ,\" kJ/kg\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.3:pg-302"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.3\n",
      "\n",
      " The enthalpy and entropy of the system are\n",
      "  2614.55463998  kW and  5.96006442363  kJ/kg and kJ/kg K respectively.\n"
     ]
    }
   ],
   "source": [
    "\n",
    "v = 0.09  # Specific volume of substance at a point in m**3/kg\n",
    "vf = 0.001177 # Specific volume of fluid in m**3/kg\n",
    "vg = 0.09963  # Specific volume of gas in m**3/kg\n",
    "hf = 908.79 # Specific enthalpy of fluid in kJ/kg\n",
    "hfg = 1890.7 # Latent heat of substance in kJ/kg\n",
    "sf = 2.4474 # Specific entropy of fluid in kJ/kgK\n",
    "sfg = 3.8935 # Entropy change due to vaporization\n",
    "\n",
    "print \"\\n Example 9.3\"\n",
    "x = (v-vf)/(vg-vf) # steam quality\n",
    "h = hf+(x*hfg) # Specific enthalpy of substance at a point in kJ/kg\n",
    "s = sf+(x*sfg) # Specific entropy of substance at a point in kJ/kgK\n",
    "\n",
    "print \"\\n The enthalpy and entropy of the system are\\n \",h ,\" kW and \",s ,\" kJ/kg and kJ/kg K respectively.\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.5:pg-303"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.5\n",
      "\n",
      " The pressure is  3.973  MPa\n",
      "\n",
      " The total mass of mixture is  9.57329343706  kg\n",
      "\n",
      " Specific volume is  0.00417829039327  m3/kg\n",
      "\n",
      " Enthalpy is is  1188.13405609  kJ/kg\n",
      "\n",
      " The entropy is  2.9891336667  kJ/kg K\n",
      "\n",
      " The internal energy is  1171.53370836  kJ/kg\n",
      "\n",
      " At 250 degree Celsius, internal energy is  1171.53445483 kJ/kg\n"
     ]
    }
   ],
   "source": [
    "\n",
    "Psat = 3.973 # Saturation pressure in MPa\n",
    "vf = 0.0012512 # specific volume of fluid in m**3/kg\n",
    "vg = 0.05013 # Specific volume of gas in m**3/kg\n",
    "hf = 1085.36  # Specific enthalpy of fluid in kJ/kg\n",
    "hfg = 1716.2 # Latent heat of vaporization in kJ/kg\n",
    "sf = 2.7927 # Specific entropy of fluid in kJ/kgK\n",
    "sfg = 3.2802 # Entropy change due to vaporization in kJ/kgK\n",
    "mf = 9.0 # Mass of liquid in kg\n",
    "V = 0.04 # Volume of vessel in m**3\n",
    "# at T = 250\n",
    "uf = 1080.39 #Specific internal energy in kJ/kg \n",
    "ufg = 1522.0# Change in internal energy due to vaporization in kJ/kg\n",
    "\n",
    "print \"\\n Example 9.5\"\n",
    "Vf = mf*vf # volume of fluid\n",
    "Vg = V-Vf # volume of gas\n",
    "mg = Vg/vg # mass of gas\n",
    "m = mf+mg # mass if mixture\n",
    "x = mg/m # quality of steam\n",
    "v = vf+x*(vg-vf) # specific volume of mixture\n",
    "h = hf+x*hfg # enthalpy of mixture\n",
    "s = sf+(x*sfg) # entropy of mixture\n",
    "u = h-Psat*1e6*v*1e-03 # Internal energy of mixture\n",
    "u_ = uf+x*ufg # Internal energy at 250 degree Celsius\n",
    "print \"\\n The pressure is \",Psat ,\" MPa\"\n",
    "print \"\\n The total mass of mixture is \",m ,\" kg\"\n",
    "print \"\\n Specific volume is \",v ,\" m3/kg\"\n",
    "print \"\\n Enthalpy is is \",h ,\" kJ/kg\"\n",
    "print \"\\n The entropy is \",s ,\" kJ/kg K\"\n",
    "print \"\\n The internal energy is \",u ,\" kJ/kg\"\n",
    "print \"\\n At 250 degree Celsius, internal energy is \",u_ ,\"kJ/kg\" #The answer provided in the textbook is wrong\n",
    "\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.7:pg-305"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.7\n",
      "\n",
      " The ideal work output of the turbine is  882.40804932  kJ/Kg\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# At T = 40 degree\n",
    "Psat = 7.384 # Saturation pressure in kPa\n",
    "sf = 0.5725 # Entropy of fluid in kJ/kgK\n",
    "sfg = 7.6845 # Entropy change due to vaporization in kJ/kgK\n",
    "hf = 167.57 # Enthalpy of fluid in kJ/kg\n",
    "hfg = 2406.7 # Latent heat of vaporization in kJ/kg\n",
    "s1 = 6.9189 # Entropy at turbine inlet in kJ/kgK\n",
    "h1 = 3037.6 # Enthalpy at turbine inlet in kJ/kg\n",
    "print \"\\n Example 9.7\"\n",
    "x2 = (s1-sf)/sfg # Steam quality\n",
    "h2 = hf+(x2*hfg) # Enthalpy at turbine exit\n",
    "W = h1-h2 # Net work done\n",
    "print \"\\n The ideal work output of the turbine is \",W ,\" kJ/Kg\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.9:pg-308"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.9\n",
      "\n",
      " The quality of steam in pipe line is  0.96097673702\n",
      "\n",
      " Maximum moisture content that can be determined is  5.47886817645  percent\n"
     ]
    }
   ],
   "source": [
    "\n",
    "h2 = 2716.2 # Enthalpy at turbine inlet in kJ/kg\n",
    "hf = 844.89 # Enthalpy of fluid in kJ/kg\n",
    "hfg = 1947.3 # Latent heat of vaporization in kJ/kg\n",
    "h3 = 2685.5 # Enthalpy at turbine exit in kJ/kg\n",
    "print \"\\n Example 9.9\"\n",
    "x1 = (h2-hf)/hfg\n",
    "x4 = (h3-hf)/hfg\n",
    "print \"\\n The quality of steam in pipe line is \",x1  #The answers vary due to round off error\n",
    "print \"\\n Maximum moisture content that can be determined is \",100-(x4*100) ,\" percent\"#The answer provided in the textbook is wrong\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.10:pg-309"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.10\n",
      "\n",
      " The quality of the steam in the pipe line is  0.909544295341\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# At 0.1Mpa, 110 degree\n",
    "h2 = 2696.2 # Enthalpy at turbine inlet in kJ/kg\n",
    "hf = 844.89 # Enthalpy of fluid in kJ/kg\n",
    "hfg = 1947.3 # Latent heat of vaporization in kJ/kg\n",
    "vf = 0.001023 # at T = 70 degree\n",
    "V = 0.000150 # In m3\n",
    "m2 = 3.24 # mass of condensed steam in kg\n",
    "\n",
    "print \"\\n Example 9.10\"\n",
    "x2 = (h2-hf)/hfg # Quality of steam at turbine inlet\n",
    "m1 = V/vf # mass of moisture collected in separator\n",
    "x1 = (x2*m2)/(m1+m2) # quality of the steam\n",
    "print \"\\n The quality of the steam in the pipe line is \",x1 \n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.11:pg-310"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.11\n",
      "\n",
      " The heat transfer during the process is  1788.19203218  MJ\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# P = 1MPa\n",
    "vf = 0.001127 # specific volume of fluid in m**3/kg\n",
    "vg = 0.1944# specific volume of gas in m**3/kg\n",
    "hg = 2778.1 # specific enthalpy of gas in kJ/kg\n",
    "uf = 761.68 # Specific internal energy of fluid in kJ/kg\n",
    "ug = 2583.6 # Specific internal energy of gas in kJ/kg\n",
    "ufg = 1822 # Change in specific internal energy due to phase change in kJ/kg \n",
    "# Initial anf final mass\n",
    "Vif = 5 # Initial volume of water in m**3 \n",
    "Viw = 5# Initial volume of gas in m**3 \n",
    "Vff = 6  # Final volume of gas in m**3 \n",
    "Vfw = 4 # Final volume of water in m**3 \n",
    "\n",
    "\n",
    "print \"\\n Example 9.11\"\n",
    "ms = ((Viw/vf)+(Vif/vg)) - ((Vfw/vf)+(Vff/vg)) \n",
    "U1 =  ((Viw*uf/vf)+(Vif*ug/vg))\n",
    "Uf =  ((Vfw*uf/vf)+(Vff*ug/vg))\n",
    "Q = Uf-U1+(ms*hg)\n",
    "print \"\\n The heat transfer during the process is \",Q/1e3 ,\" MJ\"\n",
    "#The answer provided in the textbook is wrong\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.12:pg-311"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.12\n"
     ]
    },
    {
     "ename": "NameError",
     "evalue": "name 'math' is not defined",
     "output_type": "error",
     "traceback": [
      "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
      "\u001b[1;31mNameError\u001b[0m                                 Traceback (most recent call last)",
      "\u001b[1;32m<ipython-input-5-4824fe7bf0c8>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m     14\u001b[0m \u001b[1;32mprint\u001b[0m \u001b[1;34m\"\\n Example 9.12\"\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     15\u001b[0m \u001b[0mV1\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mm\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0mv1\u001b[0m \u001b[1;31m# total volume at point 1\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 16\u001b[1;33m \u001b[0mVd\u001b[0m \u001b[1;33m=\u001b[0m \u001b[1;33m(\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mpi\u001b[0m\u001b[1;33m/\u001b[0m\u001b[1;36m4\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0md\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m1e-3\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;36m2\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0ml\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m1e-3\u001b[0m \u001b[1;31m# displaced volume\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     17\u001b[0m \u001b[0mV2\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mV1\u001b[0m\u001b[1;33m+\u001b[0m\u001b[0mVd\u001b[0m  \u001b[1;31m# Total volume at point 2\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     18\u001b[0m \u001b[0mn\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mlog\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mP1\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mP2\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mlog\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mV2\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mV1\u001b[0m\u001b[1;33m)\u001b[0m \u001b[1;31m# polytropic index\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
      "\u001b[1;31mNameError\u001b[0m: name 'math' is not defined"
     ]
    }
   ],
   "source": [
    "\n",
    "m = 0.02 # Mass of steam in Kg\n",
    "d = 280 # diameter of piston in mm\n",
    "l = 305 # Stroke length in mm\n",
    "P1 = 0.6 # Initial pressure in MPa\n",
    "P2 = 0.12 # Final pressure in MPa\n",
    "# At 0.6MPa, t = 200 degree\n",
    "v1 = 0.352 # Specific volume in m**3/kg\n",
    "h1 = 2850.1 # Specific enthalpy in kJ/kg\n",
    "vf = 0.0010476 # specific volume of fluid in m**3/kg\n",
    "vfg = 1.4271 # Specific volume change due to vaporization in m**3/kg\n",
    "uf = 439.3 # specific enthalpy of fluid\n",
    "ug = 2512.0 # Specific enthalpy of gas\n",
    "print \"\\n Example 9.12\"\n",
    "V1 = m*v1 # total volume at point 1\n",
    "Vd = (math.pi/4)*(d*1e-3)**2*l*1e-3 # displaced volume\n",
    "V2 = V1+Vd  # Total volume at point 2\n",
    "n = log(P1/P2)/log(V2/V1) # polytropic index\n",
    "W12 = ((P1*V1)-(P2*V2))*1e6/(n-1) # work done\n",
    "print \"\\n The value of n is \",n\n",
    "print \"\\n The work done by the steam is \",W12/1e3 ,\"kJ \"\n",
    "#The answers vary due to round off error\n",
    "v2 = V2/m # specific volume\n",
    "x2 = (v2-vf)/vfg  # Steam quality\n",
    "# At 0.12MPa\n",
    "u2 = uf + (x2*(ug-uf)) # Internal energy \n",
    "u1 = h1-(P1*1e6*v1*1e-03) # Internal energy\n",
    "Q12 = m*(u2-u1)+ (W12/1e3) # Heat transfer\n",
    "print \"\\n The heat transfer is \",Q12 ,\"kJ \"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.13:pg-312"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.13\n",
      "\n",
      " Final pressure is  3.5  bar\n",
      "\n",
      " Steam quality is  0.87 \n",
      " Entropy change during the process is  0.4227  kJ/K\n"
     ]
    }
   ],
   "source": [
    "\n",
    "x1 = 1 # Steam quality in first vessel\n",
    "x2 = 0.8 # Steam quality in second vessel\n",
    "# at 0.2MPa\n",
    "vg = 0.8857 # Specific volume of gas in m**3/kg\n",
    "h1 =  2706.7 # Enthalpy in first vessel  in kJ/kg\n",
    "v1 = vg # Specific volume of gas in first vessel in m**3/kg\n",
    "hg = h1 #  Enthalpy in first vessel  1 in kJ/kg\n",
    "m1 = 5  # mass in first vessel in kg\n",
    "V1 = m1*v1 # Volume of first vessel in m**3\n",
    "# at 0.5MPa\n",
    "m2 = 10 # mass in second vessel in kg\n",
    "hf = 640.23 # Enthalpy in second vessel  in kJ/kg\n",
    "hfg = 2108.5 # Latent heat of vaporization in kJ/kg\n",
    "vf = 0.001093 # Specific volume of fluid in second vessel in m**3/kg\n",
    "vfg = 0.3749 # Change in specific volume in second vessel due to evaporation of gas in m**3/kg\n",
    "v2 = vf+(x2*vfg) # Specific volume of gas in second vessel\n",
    "V2 = m2*v2 # Volume of second vessel in m**3\n",
    "#\n",
    "Vm = V1+V2 # Total volume \n",
    "m = m1+m2 # Total mass\n",
    "vm = Vm/m # net specific volume\n",
    "u1 = h1 # Internal energy\n",
    "h2 = hf+(x2*hfg) # Enthalpy calculation\n",
    "u2 = h2 # Internal energy calculation\n",
    "m3 = m # Net mass calculation\n",
    "h3 = ((m1*u1)+(m2*u2))/m3 # Resultant enthalpy calculation\n",
    "u3 = h3 # Resultant internal energy calculation\n",
    "v3 = vm # resultant specific volume calculation\n",
    "# From Mollier diagram\n",
    "x3 = 0.870  # Steam quality \n",
    "p3 = 3.5 # Pressure in MPa\n",
    "s3 = 6.29 # Entropy at state 3  in kJ/kgK\n",
    "s1 = 7.1271 # Entropy at state 1 in kJ/kgK\n",
    "sf = 1.8607  # Entropy in liquid state  in kJ/kgK\n",
    "sfg = 4.9606 # Entropy change due to vaporization in kJ/kgK\n",
    "s2 = sf+(x2*sfg) # Entropy calculation\n",
    "E = m3*s3-((m1*s1)+(m2*s2)) # Entropy change during process\n",
    "\n",
    "print \"\\n Example 9.13\"\n",
    "print \"\\n Final pressure is \",p3 ,\" bar\"\n",
    "print \"\\n Steam quality is \",x3 ,\n",
    "print \"\\n Entropy change during the process is \",E ,\" kJ/K\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.14:pg-314"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.14\n",
      "\n",
      " The availability of the steam before the throttle valve  1263.6894  kJ/kg\n",
      "\n",
      " The availability of the steam after the throttle valve  1237.5538  kJ/kg\n",
      "\n",
      " The availability of the steam at the turbine exhaust  601.851036792  kJ/kg\n",
      "\n",
      " The specific work output from the turbine is  546.253422512  kJ/kg\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# At 6 MPa, 400 degree\n",
    "h1 = 3177.2 # Enthalpy in kJ/kg\n",
    "s1 = 6.5408 #Entropy in kJ/kgK\n",
    "# At 20 degree\n",
    "h0= 83.96 # Enthalpy in kJ/kg \n",
    "s0 = 0.2966#Entropy in kJ/kgK\n",
    "T0 = 20 # Surrounding temperature in degree Celsius \n",
    "f1 = (h1-h0)-(T0+273)*(s1-s0) # Availability before throttling\n",
    "# By interpolation at P= 5MPa, h= 3177.2\n",
    "s2 = 6.63 #Entropy in kJ/kgK\n",
    "h2 = h1 # Throttling\n",
    "f2 = (h2-h0)-(T0+273)*(s2-s0) # Availability after throttling\n",
    "df = f1-f2 # Change in availability\n",
    "x3s = (s2-1.5301)/(7.1271-1.5301) #Entropy at state 3 in kJ/kgK\n",
    "h3s = 504.7+(x3s*2201.9) #Enthalpy at state 3 in kJ/kg\n",
    "eis = 0.82 # isentropic efficiency\n",
    "h3 = h2-eis*(h1-h3s) # Enthalpy at state 3 in kJ/kgK\n",
    "x3 = (h3-504.7)/2201.7 # Steam quality at state 3\n",
    "s3 = 1.5301+(x3*5.597) # Entropy at state 3\n",
    "f3 = (h3-h0)-(T0+273)*(s3-s0) # Availability at state 3\n",
    "\n",
    "print \"\\n Example 9.14\"\n",
    "print \"\\n The availability of the steam before the throttle valve \",f1 ,\" kJ/kg\"\n",
    "print \"\\n The availability of the steam after the throttle valve \",f2 ,\" kJ/kg\"\n",
    "print \"\\n The availability of the steam at the turbine exhaust \",f3 ,\" kJ/kg\"\n",
    "print \"\\n The specific work output from the turbine is \",h2-h3 ,\" kJ/kg\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.15:pg-316"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.15\n",
      "\n",
      " Availability of steam entering is  1057.4864  kJ/kg\n",
      "\n",
      " Availability of steam leaving the turbine is  656.7062  kJ/kg\n",
      "\n",
      " Maximum work is  741.14568  kJ/kg\n",
      "\n",
      " Irreversibility is  21.36505104  kJ/kg\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# At 25 bar, 350 degree\n",
    "h1 = 3125.87 # Enthalpy in kJ/kg\n",
    "s1 = 6.8481# Entropy in kJ/kgK\n",
    "# 30 degree\n",
    "h0 = 125.79 # Enthalpy in kJ/kg\n",
    "s0 = 0.4369# Entropy in kJ/kgK\n",
    "# At 3 bar, 200 degree\n",
    "h2 = 2865.5 # Enthalpy in kJ/kg\n",
    "s2 = 7.3115 #Entropy in kJ/kgK\n",
    "# At 0.2 bar 0.95 dry\n",
    "hf = 251.4 # Enthalpy of liquid in kJ/kg\n",
    "hfg = 2358.3 # Latent heat of vaporization in kJ/kg\n",
    "sf = 0.8320 # Entropy of liquid in kJ/kgK\n",
    "sg = 7.0765# Entropy of liquid in kJ/kgK\n",
    "h3 = hf+0.92*hfg # Enthalpy at state 3 in kJ/kg\n",
    "s3 = sf+(0.92*sg) # Entropy at state 3 in kJ/kgK\n",
    "# Part (a)\n",
    "T0 = 30 # Atmospheric temperature in degree Celsius\n",
    "f1 = (h1-h0)-((T0+273)*(s1-s0)) # Availability at steam entering turbine\n",
    "f2 = (h2-h0)-((T0+273)*(s2-s0)) # Availability at state 2\n",
    "f3 = (h3-h0)-((T0+273)*(s3-s0))# Availability at state 3\n",
    "\n",
    "print \"\\n Example 9.15\"\n",
    "print \"\\n Availability of steam entering is \",f1 ,\" kJ/kg\"\n",
    "print \"\\n Availability of steam leaving the turbine is \",f2 ,\" kJ/kg\"\n",
    "\n",
    "# Part (b)\n",
    "m2m1 = 0.25  # mass ratio\n",
    "m3m1 = 0.75 # mass ratio\n",
    "Wrev = f1-(m2m1*f2)-(m3m1*f3) # Maximum work\n",
    "print \"\\n Maximum work is \",Wrev ,\" kJ/kg\"\n",
    "\n",
    "# Part (c)\n",
    "w1 = 600 # mass flow at inlet of turbine in kg/h\n",
    "w2 = 150 # mass flow at state 2 in turbine in kg/h\n",
    "w3 = 450# mass flow at state 2 in  turbine in kg/h\n",
    "Q = -10  # Heat loss rate kJ/s\n",
    "I = ((T0+273)*(w2*s2+w3*s3-w1*s1)-Q*3600)*103/600\n",
    "print \"\\n Irreversibility is \",I/1e3 ,\" kJ/kg\"\n",
    "#The answer provided in the textbook is wrong\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.16:pg-317"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.16\n",
      "\n",
      " Energy of system in Part (a) is  73.2  kJ\n",
      "\n",
      " Energy of system in Part (b) is  197.3474  kJ\n",
      "\n",
      " Energy of system in Part (c) is  498.2624  kJ\n",
      "\n",
      " Energy of system in Part (d) is  121.8  kJ\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# At dead state of 1 bar, 300K\n",
    "u0 = 113.1 # Internal energy in kJ/kg\n",
    "h0 = 113.2 # Enthalpy in kJ/kg\n",
    "v0 = 0.001005 # Specific volume in m**3/kg\n",
    "s0 = 0.395 # Entropy in kJ/kg\n",
    "T0 = 300 # Atmospheric temperature in K\n",
    "P0 = 1 # Atmospheric pressure in bar \n",
    "K = h0-T0*s0\n",
    "# Part (a)\n",
    "# At 1bar and 90 degree Celsius \n",
    "u = 376.9 # Internal energy in kJ/kg\n",
    "h = 377 # Enthalpy in kJ/kg\n",
    "v = 0.001035 # specific volume in m**3/kg\n",
    "s = 1.193 # Entropy in kJ/kgK\n",
    "m = 3 # Mass of water in kg\n",
    "fi = m*(h-(T0*s)-K)  #Energy of system\n",
    "\n",
    "print \"\\n Example 9.16\"\n",
    "print \"\\n Energy of system in Part (a) is \",fi ,\" kJ\"\n",
    "#The answers vary due to round off error\n",
    "\n",
    "# Part (b)\n",
    "# At P = 4 Mpa, t = 500 degree\n",
    "u = 3099.8# Internal energy in kJ/kg \n",
    "h = 3446.3 # Enthalpy in kJ/kg \n",
    "v = 0.08637 # specific volume in m**3/kg \n",
    "s = 7.090 # Entropy in kJ/kgK\n",
    "m = 0.2 # Mass of steam in kg \n",
    "fib = m*(u+P0*100*v-T0*s-K) # Energy of system\n",
    "print \"\\n Energy of system in Part (b) is \",fib ,\" kJ\"\n",
    "\n",
    "# Part (c) # P = 0.1 bar\n",
    "m = 0.4 # Mass of wet steam in kg \n",
    "x = 0.85 # Quality\n",
    "u = 192+x*2245 # Internal energy \n",
    "h = 192+x*2392# Enthalpy\n",
    "s = 0.649+x*7.499 # Entropy\n",
    "v = 0.001010+x*14.67 # specific volume\n",
    "fic = m*(u+P0*100*v-T0*s-K) # Energy of system\n",
    "print \"\\n Energy of system in Part (c) is \",fic ,\" kJ\"\n",
    "\n",
    "# Part (d) \n",
    "# P = 1 Bar, t = -10 degree Celsius\n",
    "m = 3 # Mass of ice in kg \n",
    "h = -354.1 # Enthalpy in kJ/kg  \n",
    "s = -1.298 # at 1000kPa, -10 degree\n",
    "fid = m*((h-h0)-T0*(s-s0)) # Energy of system\n",
    "\n",
    "print \"\\n Energy of system in Part (d) is \",fid ,\" kJ\" #The answer provided in the textbook is wrong\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.17:pg-318"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.17\n",
      "\n",
      " In parallel flow\n",
      "\n",
      " The rate of irreversibility is  10.98  kW\n",
      "\n",
      " The Second law efficiency is  24.275862069  percent\n",
      "\n",
      "\n",
      " In counter flow\n",
      "\n",
      " The rate of irreversibility is  10.9454545455  kW\n",
      "\n",
      " The Second law efficiency is  32.1594684385  percent\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# Given\n",
    "th1 = 90.0 # Inlet temperature of hot water in degree Celsius\n",
    "tc1 = 25.0# Inlet temperature of cold water in degree Celsius\n",
    "tc2 = 50.0# Exit temperature of cold water in degree Celsius\n",
    "mc = 1.0 # mass flow rate of cold water in kg/s\n",
    "T0 = 300.0 # Atmospheric temperature in K\n",
    "th2p = 60.0 # Temperature limit in degree Celsius for parallel flow\n",
    "th2c = 35.0 # Temperature limit in degree Celsius for counter flow\n",
    "mhp = (tc2-tc1)/(th1-th2p) # mass flow rate of hot water in kg/s for parallel flow\n",
    "mhc = (tc2-tc1)/(th1-th2c) # mass flow rate of hot water in kg/s for counter flow\n",
    "# At 300 K\n",
    "h0 = 113.2 # ENthalpy in kJ/kg\n",
    "s0 = 0.395 # ENtropy in kJ/kgK\n",
    "T0 = 300.0 # temperature in K\n",
    "# At 90 degree celsius\n",
    "h1 = 376.92 # Enthalpy in kJ/kg \n",
    "s1 = 1.1925 # Entropy in kJ/kgK\n",
    "af1 = mhp*((h1-h0)-T0*(s1-s0)) # Availability\n",
    "# Parallel Flow\n",
    "# At 60 degree\n",
    "h2 = 251.13 # Enthalpy in kJ/kg \n",
    "s2 =0.8312  # Entropy in kJ/kgK\n",
    " # At 25 degree\n",
    "h3 = 104.89 # Enthalpy in kJ/kg \n",
    "s3 = 0.3674 # Entropy in kJ/kgK\n",
    "# At 50 degree\n",
    "h4 = 209.33 # Enthalpy in kJ/kg \n",
    "s4 = 0.7038 # Entropy in kJ/kgK\n",
    "REG = mc*((h4-h3)-T0*(s4-s3)) # Rate of energy gain\n",
    "REL = mhp*((h1-h2)-T0*(s1-s2)) # Rate of energy loss\n",
    "Ia = REL-REG # Energy destruction\n",
    "n2a = REG/REL # Second law efficiency\n",
    "\n",
    "print \"\\n Example 9.17\"\n",
    "print \"\\n In parallel flow\"\n",
    "print \"\\n The rate of irreversibility is \",Ia ,\" kW\"\n",
    "print \"\\n The Second law efficiency is \",n2a*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n",
    "\n",
    "\n",
    "# Counter flow\n",
    "h2_ = 146.68 \n",
    "sp = 0.5053 # At 35 degree\n",
    "REG_b = REG # Rate of energy gain by hot water is same in both flows\n",
    "REL_b = mhc*((h1-h2_)-T0*(s1-sp))\n",
    "Ib = mhc*((h1-h2_)-(T0*(s1-sp))) # Energy destruction\n",
    "n2b = REG_b/Ib # Second law efficiency\n",
    "print \"\\n\\n In counter flow\"\n",
    "print \"\\n The rate of irreversibility is \",Ib ,\" kW\"\n",
    "print \"\\n The Second law efficiency is \",n2b*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex9.18:pg-320"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 9.18\n",
      "\n",
      " The maximum cooling rate is  106.207042424  kW\n"
     ]
    }
   ],
   "source": [
    "\n",
    "m = 50.0# mass flow rate in kg/h\n",
    "Th = 23.0 # Home temperature in degree Celsius\n",
    "# State 1\n",
    "T1 = 150.0 # Saturated vapor temperature in degree Celsius\n",
    "h1 = 2746.4 # Saturated vapor enthalpy in kJ/kg\n",
    "s1 = 6.8387 #Saturated vapor entropy in kJ/kgK\n",
    "# State 2\n",
    "h2 = 419.0 # Saturated liquid enthalpy in kJ/kg\n",
    "s2 = 1.3071 #Saturated liquid entropy in kJ/kg \n",
    "T0 = 45.0 #  Atmospheric temperature in degree Celsius\n",
    "#\n",
    "b1 = h1-((T0+273)*s1) # Availability at point 1\n",
    "b2 = h2-((T0+273)*s2) # Availability at point 2\n",
    "Q_max = m*(b1-b2)/((T0+273)/(Th+273)-1) # maximum cooling rate\n",
    "\n",
    "print \"\\n Example 9.18\"\n",
    "print \"\\n The maximum cooling rate is \",Q_max/3600 ,\" kW\"\n",
    "\n"
   ]
  }
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