{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 08: Available energy Availability and irreversibility"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.1:pg-249"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.1\n",
      "\n",
      " The fraction of energy that becomes unavailable due to irreversible heat transfer is  0.260038240918\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "T0 = 35.0 # Heat rejection temperature in degree Celsius \n",
    "T1 = 420 #  Vapor condensation temperature in degree Celsius \n",
    "T1_ = 250 # water vapor temperature in degree Celsius \n",
    "print \"\\n Example 8.1\"\n",
    "f = ((T0+273)*((T1+273)-(T1_+273)))/((T1_+273)*((T1+273)-(T0+273)))# fraction of energy lost\n",
    "print \"\\n The fraction of energy that becomes unavailable due to irreversible heat transfer is \",f \n",
    "#The answers vary due to round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.2:pg-250"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.2\n",
      "\n",
      " Total change in entropy is  2.03990232306  kJ/K\n",
      "\n",
      " Increase in unavailable energy is  618.090403887  kJ\n"
     ]
    }
   ],
   "source": [
    "from scipy import integrate\n",
    "import math\n",
    "\n",
    "lhw = 1858.5 # Latent heat of water in kJ/kg\n",
    "Tew = 220 # Water evaporation temperature in degree Celsius\n",
    " \n",
    "Tig = 1100 # Initial temperature of the gas in degree Celsius\n",
    "Tfg = 550 # Final temperature of the gas in degree Celsius\n",
    "T0 = 303 # Atmospheric temperature in degree Celsius\n",
    "Tg2 = 823 \n",
    "Tg1 = 1373\n",
    "print \"\\n Example 8.2\"\n",
    "Sw = lhw/(Tew+273) # Entropy generation in water\n",
    "Sg,error = integrate.quad(lambda T:3.38/T,Tg1,Tg2)\n",
    "St = Sg+Sw \n",
    "print \"\\n Total change in entropy is \",St ,\" kJ/K\"\n",
    "\n",
    "print \"\\n Increase in unavailable energy is \",T0*St ,\" kJ\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.4:pg-253"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.4\n",
      "\n",
      " The decrease in the available energy is  281.816890623  kJ\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "from scipy import integrate\n",
    "Ts_ = 15 # Ambient temperature in degree Celsius\n",
    "Tw1_ = 95 # Temperature of water sample 1 in degree Celsius\n",
    "Tw2_ = 35# Temperature of water sample 2 in degree Celsius\n",
    "m1 = 25 # Mass of water sample 1 in kg\n",
    "m2 = 35 # Mass of water sample 2 in kg\n",
    "cp = 4.2 # Specific heat capacity of water in kJ/kgK\n",
    "print \"\\n Example 8.4\"\n",
    "Ts = Ts_+273# Ambient temperature in K\n",
    "Tw1 = Tw1_+273 # Temperature of water sample 1 in K\n",
    "Tw2 = Tw2_+273# Temperature of water sample 2 in K\n",
    "AE25,er = integrate.quad(lambda T:m1*cp*(1-(Ts/T)),Ts,Tw1)\n",
    "AE35,er2 = integrate.quad(lambda T:m2*cp*(1-(Ts/T)),Ts,Tw2)\n",
    "AEt = AE25 + AE35\n",
    "Tm = (m1*Tw1+m2*Tw2)/(m1+m2) # Temperature after mixing\n",
    "AE60,er3 = integrate.quad(lambda T:(m1+m2)*cp*(1-(Ts/T)),Ts,Tm)\n",
    "AE = AEt - AE60\n",
    "print \"\\n The decrease in the available energy is \",AE ,\" kJ\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.5:pg-254"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.5\n",
      "\n",
      " The final RPM of the flywheel would be  222.168786807  RPM\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "from scipy import integrate\n",
    "N1 = 3000 # Speed of rotation of flywheel in RPM\n",
    "I = 0.54 # Moment of inertia of flywheel in kgm**2\n",
    "ti_ = 15 # Temperature of insulated system in degree Celsius \n",
    "m = 2 # Water equivalent of shaft \n",
    "print \"\\n Example 8.5\"\n",
    "w1 = (2*math.pi*N1)/60 # Angular velocity of rotation in rad/s\n",
    "Ei = 0.5*I*w1**2 # rotational kinetic energy\n",
    "dt = Ei/(1000*2*4.187) # temperature change\n",
    "ti = ti_+273# Temperature of insulated system in Kelvin\n",
    "tf = ti+dt # final temperature\n",
    "AE,er = integrate.quad(lambda T: m*4.187*(1-(ti/T)),ti,tf)\n",
    "UE = Ei/1000 - AE # Unavailable enrgy\n",
    "w2 = math.sqrt(AE*1000*2/I) # Angular speed in rad/s \n",
    "N2 = (w2*60)/(2*math.pi) # Speed of rotation in RPM\n",
    "print \"\\n The final RPM of the flywheel would be \",N2 ,\" RPM\"\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.6:pg-255"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.6\n",
      "\n",
      " The maximum work is  122.957271378  kJ\n",
      "\n",
      " Change in availability is  82.4328713783  kJ\n",
      "\n",
      " Irreversibility is  15.2572713783  kJ\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "from scipy import integrate\n",
    "T1_ = 80.0 # Initial temperature of air in degree Celsius \n",
    "T2_ = 5.0 # Final temperature of air in degree Celsius \n",
    "V2 = 2.0 # Assumed final volume\n",
    "V1 = 1.0 # Assumed initial volume\n",
    "P0 = 100.0 # Final pressure of air in kPa\n",
    "P1 = 500.0 # Initial pressure of air in kPa\n",
    "R = 0.287 # Gas constant\n",
    "cv = 0.718 # Specific heat capacity at constant volume for gas in kJ/kg K\n",
    "m = 2.0 # Mass of gas in kg\n",
    "print \"\\n Example 8.6\"\n",
    "T1= T1_+273 # Initial temperature of air in K \n",
    "T2 = T2_+273 # Final temperature of air in K \n",
    "S= integrate.quad(lambda T:(m*cv)/T,T1,T2)[0] + integrate.quad(lambda V: (m*R)/V,V1,V2)[0] # Entropy change\n",
    "U = m*cv*(T1-T2)# Change in internal energy\n",
    "Wmax = U-(T2*(-S)) # Maximum possible work\n",
    "V1_ = (m*R*T1)/P1 # volume calculation\n",
    "CA = Wmax-P0*(V1_) # Change in availability\n",
    "I = T2*S # Irreversibility\n",
    "print \"\\n The maximum work is \",Wmax ,\" kJ\"\n",
    "print \"\\n Change in availability is \",CA ,\" kJ\"\n",
    "print \"\\n Irreversibility is \",I ,\" kJ\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.7:pg-256"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.7\n",
      "\n",
      " The decrease in availability is  260.756521108  kJ/kg\n",
      "\n",
      " The maximum work is  260.756521108  kJ/kg\n",
      "\n",
      " The irreversibility is  49.6565211082  kJ/kg\n",
      "\n",
      " Alternatively, The irreversibility is  49.6565211082  kJ/kg\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "P1 = 500.0 # Initial pressure of steam in kPa\n",
    "P2 = 100.0# Final pressure of steam in kPa\n",
    "T1_ = 520.0 #Initial temperature of steam in degree Celsius\n",
    "T2_ = 300.0 #Final temperature of steam in degree Celsius\n",
    "cp = 1.005 # Specific heat capacity of steam in kJ/kgK\n",
    "t0 = 20.0 # Atmospheric temperature in degree Celsius \n",
    "R = 0.287 # Gas constant\n",
    "Q = -10.0 # Heat loss to surrounding in kJ/kg\n",
    "print \"\\n Example 8.7\"\n",
    "T1 = T1_+273 #Initial temperature of steam in degree Celsius\n",
    "T2 = T2_+273 #Final temperature of steam in degree Celsius\n",
    "S21 = (R*math.log(P2/P1))-(cp*math.log(T2/T1))\n",
    "T0 = t0+273\n",
    "CA = cp*(T1-T2)-T0*S21 # Change in availability\n",
    "Wmax = CA # Maximum possible work\n",
    "W = cp*(T1-T2)+Q # net work\n",
    "I = Wmax-W # Irreversibility\n",
    "# Altenatively\n",
    "Ssystem = -Q/T0\n",
    "Ssurr = -S21\n",
    "I1 = T0*(Ssystem+Ssurr)\n",
    "print \"\\n The decrease in availability is \",CA ,\" kJ/kg\"\n",
    "print \"\\n The maximum work is \",Wmax ,\" kJ/kg\"\n",
    "print \"\\n The irreversibility is \",I ,\" kJ/kg\"\n",
    "print \"\\n Alternatively, The irreversibility is \",I1 ,\" kJ/kg\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.8:pg-258"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.8\n",
      "\n",
      " The initial and final availbility of the products are  85.9672398469  kJ/Kg and  39.6826771757  kJ/Kg respectively\n",
      "\n",
      " The irreversibility of the process is  319.369801955  kW\n",
      "\n",
      " Total power generated by the heat engine is  472.671938045  kW\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "T0 = 300.0 # Atmospheric temperature in K\n",
    "Tg1_ = 300.0 # Higher temperature of combustion product in degree Celcius\n",
    "Tg2_ = 200.0 # Lower temperature of combustion product in degree Celcius\n",
    "Ta1 = 40.0 # Initial air temperature in K\n",
    "cpg = 1.09 # Specific heat capacity of combustion gas in kJ/kgK\n",
    "cpa = 1.005# Specific heat capacity of air in kJ/kgK\n",
    "mg = 12.5 # mass flow rate of product in kg/s\n",
    "ma = 11.15# mass flow rate of air in kg/s\n",
    "\n",
    "print \"\\n Example 8.8\"\n",
    "Tg1 = Tg1_+273 # Higher temperature of combustion product in K\n",
    "Tg2 = Tg2_+273 # Lower temperature of combustion product in K\n",
    "f1 = cpg*(Tg1-T0)-T0*cpg*(math.log(Tg1/T0)) # Initial availability of product\n",
    "f2 = cpg*(Tg2-T0)-T0*cpg*(math.log(Tg2/T0)) # Final availabilty of product\n",
    "print \"\\n The initial and final availbility of the products are \",f1 ,\" kJ/Kg and \",f2 ,\" kJ/Kg respectively\"\n",
    "#The answer provided in the textbook is wrong\n",
    "\n",
    "# Part (b)\n",
    "Dfg = f1-f2 # Decrease in availability of products\n",
    "Ta2 = (Ta1+273) + (mg/ma)*(cpg/cpa)*(Tg1-Tg2) # Exit temperature of air\n",
    "Ifa = cpa*(Ta2-(Ta1+273))-T0*cpa*(math.log(Ta2/(Ta1+273))) # Increase in availability of air\n",
    "I = mg*Dfg-ma*Ifa # Irreversibility \n",
    "print \"\\n The irreversibility of the process is \",I ,\" kW\"\n",
    "##The answer provided in the textbook contains round off error\n",
    "\n",
    "# Part (c)\n",
    "Ta2_ = (Ta1+273)*(Tg1/Tg2)**((12.5*1.09)/(11.5*1.005))\n",
    "Q1 = mg*cpg*(Tg1-Tg2) # Heat supply rate from gas to working fluid\n",
    "Q2 = ma*cpa*(Ta2_-(Ta1+273))# Heat rejection rate from the working fluid in heat engine\n",
    "W = Q1-Q2 # Power developed by heat engine\n",
    "print \"\\n Total power generated by the heat engine is \",W ,\" kW\"\n",
    "#The answer provided in the textbook contains round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.9:pg-260"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.9\n",
      "\n",
      " The irreversibility rate is  15.8201795694  kW\n",
      "\n",
      " The irreversibility rate at lower temperature is  3.03317755354  kW\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "T2 = 790.0 # Final temperature of gas in degree Celsius\n",
    "T1 = 800.0 # Initial temperature of gas in degree Celsius\n",
    "m = 2.0 # Mass flow rate in kg/s\n",
    "cp = 1.1 # Specific heat capacity in kJ/KgK\n",
    "T0 = 300.0 # Ambient temperature in K\n",
    "\n",
    "print \"\\n Example 8.9\"\n",
    "I = m*cp*(((T1+273)-(T2+273))-T0*(math.log((T1+273)/(T2+273)))) # irreversibility rate\n",
    "print \"\\n The irreversibility rate is \",I ,\" kW\"\n",
    "\n",
    "# At lower temperature\n",
    "T1_ = 80.0 # Initial temperature of gas in degree Celsius\n",
    "T2_ = 70.0 # Initial temperature of gas in degree Celsius\n",
    "I_ = m*cp*(((T1_+273)-(T2_+273))-T0*(math.log((T1_+273)/(T2_+273)))) # irreversibility rate\n",
    "print \"\\n The irreversibility rate at lower temperature is \",I_ ,\" kW\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.10:pg-261"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.10\n",
      "\n",
      " The rate of energy loss because of the pressure drop due to friction  25.83  kW\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "m = 3 # Mass flow rate in kg/s\n",
    "R = 0.287 # Gas constant\n",
    "T0 = 300 # Ambient temperature in K\n",
    "k = 0.10 # Fractional pressure drop\n",
    "print \"\\n Example 8.10\"\n",
    "Sgen = m*R*k # Entropy generation\n",
    "I = Sgen*T0 # Irreversibility Calculation\n",
    "print \"\\n The rate of energy loss because of the pressure drop due to friction \",I ,\" kW\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.11:pg-261"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.11\n",
      "\n",
      " The rate of entropy generation is  0.0446035560498  kW/K\n",
      "\n",
      " The rate of energy loss due to mixing is  13.3810668149  kW\n",
      "\n",
      " The rate of energy loss due to mixing is  13.3810668149  kW\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "m1 = 2.0 # Flow rate of water in kg/s\n",
    "m2 = 1.0 # Flow rate of another stream in kg/s\n",
    "T1 = 90.0 # Temperature of water in degree Celsius\n",
    "T2 = 30.0# Temperature of another stream in degree Celsius\n",
    "T0 =300.0 # Ambient temperature in K\n",
    "cp = 4.187 # Specific heat capacity of water in kJ/kgK\n",
    "\n",
    "print \"\\n Example 8.11\"\n",
    "m = m1+m2 # Net mass flow rate\n",
    "x = m1/m # mass fraction\n",
    "t = (T2+273)/(T1+273) # Temperature ratio\n",
    "Sgen = m*cp*math.log((x+t*(1-x))/(t**(1-x))) # Entropy generation\n",
    "I = T0*Sgen # Irreversibility production\n",
    "# Alternatively\n",
    "T  = (m1*T1+m2*T2)/(m1+m2) # equilibrium temperature\n",
    "Sgen1 = m1*cp*math.log((T+273)/(T1+273))+m2*cp*math.log((T+273)/(T2+273))# Entropy generation\n",
    "I1 = T0*Sgen1 # Irreversibility production\n",
    "print \"\\n The rate of entropy generation is \",Sgen ,\" kW/K\"\n",
    "print \"\\n The rate of energy loss due to mixing is \",I ,\" kW\"\n",
    "print \"\\n The rate of energy loss due to mixing is \",I1 ,\" kW\" # Calculation from alternative way\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.12:pg-262"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.12\n",
      " \n",
      "\n",
      " PART (A)\n",
      "\n",
      " The first law efficiency is  96.0  percent\n",
      "\n",
      " The second law efficiency is  79.0588235294  percent\n",
      " \n",
      "\n",
      " PART (B)\n",
      "\n",
      " The first law efficiency is  90.0  percent\n",
      "\n",
      " The second law efficiency is  42.3529411765  percent\n",
      " \n",
      "\n",
      " PART (C)\n",
      "\n",
      " The first law efficiency is  60.0  percent\n",
      "\n",
      " The second law efficiency is  4.41176470588   percent\n",
      " \n",
      "\n",
      " PART (D)\n",
      "\n",
      " The First law efficiency for all the three cases would remain same and here is  90.0  percent\n",
      "\n",
      " The Second law efficiency of part (a) is  74.1176470588  percent\n",
      "\n",
      " The Second law efficiency of part (b) is  42.3529411765  percent\n",
      "\n",
      " The Second law efficiency of part (c) is  6.61764705882  percent\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "Qr = 500.0 # Heat release in kW\n",
    "Tr = 2000.0 # Fuel burning temperature in K \n",
    "T0 = 300.0 # Ambient temperature in K\n",
    "# Part (a)\n",
    "print \"\\n Example 8.12\"\n",
    "Qa = 480.0 # Energy absorption by furnace in kW\n",
    "Ta = 1000.0 # Furnace temperature in K \n",
    "n1a = (Qa/Qr) # first law efficiency\n",
    "n2a = n1a*(1.0-(T0/Ta))/(1.0-(T0/Tr)) #second law efficiency\n",
    "\n",
    "#The answers vary due to round off error\n",
    "print \" \\n\\n PART (A)\"\n",
    "print \"\\n The first law efficiency is \",n1a*100 ,\" percent\" \n",
    "print \"\\n The second law efficiency is \",n2a*100 ,\" percent\"\n",
    "\n",
    "# Part (b)\n",
    "Qb = 450.0 # Energy absorption in steam generation in kW\n",
    "Tb = 500.0# steam generation temperature in K \n",
    "n1b = (Qb/Qr)# first law efficiency\n",
    "n2b = n1b*(1.0-(T0/Tb))/(1.0-(T0/Tr))#second law efficiency\n",
    "print \" \\n\\n PART (B)\"\n",
    "print \"\\n The first law efficiency is \",n1b*100 ,\" percent\" \n",
    "print \"\\n The second law efficiency is \",n2b*100 ,\" percent\"\n",
    "# Part (c)\n",
    "Qc = 300.0 # Energy absorption in chemical process in kW\n",
    "Tc = 320.0 # chemical process temperature in K \n",
    "n1c = (Qc/Qr) # first law efficiency\n",
    "n2c = n1c*(1.0-(T0/Tc))/(1.0-(T0/Tr))#second law efficiency\n",
    "print \" \\n\\n PART (C)\"\n",
    "print \"\\n The first law efficiency is \",n1c*100 ,\" percent\"\n",
    "print \"\\n The second law efficiency is \",n2c*100 ,\"  percent\" \n",
    "# Part (d)\n",
    "Qd = 450.0 \n",
    "n1d = (Qd/Qr)\n",
    "n2a_= n1d*(1.0-(T0/Ta))/(1.0-(T0/Tr))\n",
    "n2b_= n1d*(1.0-(T0/Tb))/(1.0-(T0/Tr))\n",
    "n2c_= n1d*(1.0-(T0/Tc))/(1.0-(T0/Tr))\n",
    "print \" \\n\\n PART (D)\"\n",
    "print \"\\n The First law efficiency for all the three cases would remain same and here is \",n1d*100 ,\" percent\" #The answer provided in the textbook is wrong\n",
    "\n",
    "print \"\\n The Second law efficiency of part (a) is \",n2a_*100 ,\" percent\"\n",
    "\n",
    "print \"\\n The Second law efficiency of part (b) is \",n2b_*100 ,\" percent\"\n",
    "\n",
    "print \"\\n The Second law efficiency of part (c) is \",n2c_*100 ,\" percent\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.14:pg-265"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.14\n",
      "\n",
      " The power input is  -235.675  kW\n",
      " \n",
      " The second law efficiency of the compressor is  85.5494233193  percent\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "cp = 1.005 # Specific heat capacity of air in kJ/kgK \n",
    "T2 = 160.0 # Compressed air temperature in degree Celsius\n",
    "T1 = 25.0 # Ambient temperature\n",
    "T0 = 25.0 # Ambient temperature\n",
    "R = 0.287 # Gas constant\n",
    "P2 = 8.0 # Pressure ratio\n",
    "P1 = 1.0 # Initial pressure of gas in bar\n",
    "Q = -100.0 # Heat loss to surrounding in kW\n",
    "m = 1.0 # Mass flow rate in kg/s\n",
    "\n",
    "print \"\\n Example 8.14\"\n",
    "W = Q + m*cp*((T1+273)-(T2+273)) # power input\n",
    "AF = cp*((T2+273)- (T1+273))-(T0+273)*((cp*math.log((T2+273)/(T1+273))-(R*math.log(P2/P1))))  # Availability\n",
    "e = AF/-W # efficiency \n",
    "print \"\\n The power input is \",W ,\" kW\"\n",
    "print \" \\n The second law efficiency of the compressor is \",e*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.15:pg-265"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.15\n",
      "\n",
      " The exergy of the complete vacuum is  100.0  kJ\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Since vacuum has zero mass\n",
    "U = 0 # Initial internal energy in kJ/kg\n",
    "H0 = 0 # Initial enthalpy in kJ/kg\n",
    "S = 0 # Initial entropy in kJ/kgK\n",
    "# If the vacuum has reduced to dead state\n",
    "U0 = 0 # Final internal energy in kJ/kg\n",
    "H0 = 0  # Final enthalpy in kJ/kg\n",
    "S0 = 0 # Final entropy in kJ/kgK\n",
    "V0 = 0 # Final volume in m**3\n",
    "P0 = 1.0 # Pressure in bar\n",
    "V = 1.0 # Volume of space in m**3\n",
    "fi = P0*1e5*V\n",
    "\n",
    "print \"\\n Example 8.15\"\n",
    "print \"\\n The exergy of the complete vacuum is \",fi/1e3 ,\" kJ\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.16:pg-266"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.16\n",
      "\n",
      " Exergy produced is  34.6210270729  MJ or  9.61695196469  kWh\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "m = 1000.0 # Mass of fish in kg \n",
    "T0 = 300.0 # Ambient temperature in K\n",
    "P0 = 1.0 # Ambient pressure in bar\n",
    "T1 = 300.0 # Initial temperature of fish in K\n",
    "T2_ = -20.0 # Final temperature of fish in degree Celsius\n",
    "Tf_ = -2.2 # Freezing point temperature of fish in degree Celsius\n",
    "Cb = 1.7 # Specific heat of fish below freezing point in kJ/kg\n",
    "Ca = 3.2 # Specific heat of fish above freezing point in kJ/kg\n",
    "Lh = 235.0 # Latent heat of fusion of fish in kJ/kg \n",
    "\n",
    "print \"\\n Example 8.16\"\n",
    "T2 = T2_+273 # Final temperature of fish in K\n",
    "Tf = Tf_+273 # Freezing point temperature of fish in K\n",
    "H12 = m*((Cb*(Tf-T2))+Lh+(Ca*(T1-Tf))) # Enthalpy change \n",
    "H21 = -H12 # Enthalpy change \n",
    "S12 = m*((Cb*math.log(Tf/T2))+(Lh/Tf)+(Ca*math.log(T1/Tf))) # Entropy change\n",
    "S21 = -S12 # Entropy change\n",
    "E = H21-T0*S21 #Exergy produced\n",
    "print \"\\n Exergy produced is \",E/1e3 ,\" MJ or \",E/3600 ,\" kWh\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.17:pg-267"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.17\n",
      "\n",
      " The irreversibility in case a is  110.031839359  kJ/kg\n",
      "\n",
      " The irreversibility in case b is  38.2318393592  kJ/kg\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "cv = 0.718 # Specific heat capacity of air in kJ/kg\n",
    "T2 = 500.0 # Final temperature of air in K\n",
    "T1 = 300.0# Initial temperature of air in K\n",
    "m = 1.0 # Mass of air in kg\n",
    "T0 = 300.0 # Ambient temperature\n",
    "# Case (a)\n",
    "print \"\\n Example 8.17\"\n",
    "Sua = cv*math.log(T2/T1) # Entropy change of universe\n",
    "Ia = T0*Sua # irreversibility\n",
    "print \"\\n The irreversibility in case a is \",Ia ,\" kJ/kg\"\n",
    "\n",
    "# Case (b)\n",
    "Q = m*cv*(T2-T1) # Heat transfer\n",
    "T = 600 # Temperature of thermal reservoir in K\n",
    "Sub = Sua-(Q/T) # Entropy change of universe\n",
    "Ib = T0*Sub # irreversibility\n",
    "print \"\\n The irreversibility in case b is \",Ib ,\" kJ/kg\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.18:pg-268"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.18\n",
      "\n",
      " Irreversibility per unit mass is  142.7096  kJ/kg\n",
      "\n",
      " The second law efficiency of the turbine is  78.0527289547  percent\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "h1 = 3230.9 # Enthalpy of steam at turbine inlet in kJ/kg\n",
    "s1 = 6.69212# Entropy of steam at turbine inlet in kJ/kgK \n",
    "V1 = 160.0 # Velocity of steam at turbine inlet in m/s\n",
    "T1 = 400.0 # Temperature of steam at turbine inlet in degree Celsius\n",
    "h2 = 2676.1 # Enthalpy of steam at turbine exit in kJ/kg\n",
    "s2 = 7.3549 # Entropy of steam at turbine exit in kJ/kgK \n",
    "V2 = 100.0 # Velocity of steam at turbine exit in m/s\n",
    "T2 = 100.0 # Temperature of steam at turbine exit in degree Celsius\n",
    "T0 = 298.0 # Ambient temperature in K\n",
    "W = 540.0 # Work developed by turbine in kW\n",
    "Tb = 500.0 # Average outer surface temperature of turbine in K\n",
    "\n",
    "print \"\\n Example 8.18\"\n",
    "Q = (h1-h2)+((V1**2-V2**2)/2)*1e-03-W # Heat loss\n",
    "I = 151.84-Q*(0.404) # Irreversibility \n",
    "AF = W + Q*(1.0-(T0/Tb)) + I # Exergy transfer\n",
    "n2 = W/AF # second law efficiency\n",
    "\n",
    "print \"\\n Irreversibility per unit mass is \",I ,\" kJ/kg\"\n",
    "print \"\\n The second law efficiency of the turbine is \",n2*100 ,\" percent\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.19:pg-269"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.19\n",
      "\n",
      " Case A:\n",
      "\n",
      " Rate of availability transfer with heat and the irreversibility rate are \n",
      "   1.7  kW and  -6.8  kW respectively.\n",
      "\n",
      " Case B:\n",
      "\n",
      " Rate of availability in case b is  3.4  kW \n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "T0 = 300.0 # Ambient temperature in K\n",
    "T = 1500.0 # Resistor temperature in K\n",
    "Q = -8.5 # Power supply in kW\n",
    " \n",
    "# Case (a)\n",
    "W = -Q # work transfer\n",
    "I = Q*(1.0-T0/T) + W # Irreversibility\n",
    "R = Q*(1.0-T0/T) # availability\n",
    "\n",
    "print \"\\n Example 8.19\"\n",
    "print \"\\n Case A:\"\n",
    "print \"\\n Rate of availability transfer with heat and the irreversibility rate are \\n  \",I ,\" kW and \",R ,\" kW respectively.\"\n",
    "# Case (b)\n",
    "T1 = 500.0 # Furnace wall temperature\n",
    "Ib = - Q*(1.0-T0/T) + Q*(1.0-T0/T1) # Irreversibility\n",
    "print \"\\n Case B:\"\n",
    "print \"\\n Rate of availability in case b is \",Ib ,\" kW \"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.20:pg-270"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 8.20\n",
      "\n",
      "\n",
      " Part A:\n",
      "\n",
      " There is heat loss to surrounding.\n",
      "\n",
      "\n",
      " Part B:\n",
      "\n",
      " The polytropic index is  1.0\n",
      "\n",
      "\n",
      " Part C:\n",
      "\n",
      " Isothermal efficiency is  97.8793558312  percent \n",
      "\n",
      "\n",
      " Part D:\n",
      "\n",
      " The minimum work input is  -6.44697949667  kJ/kg, and irreversibility is  108.941520503  kJ/kg\n",
      "\n",
      "\n",
      " Part E:\n",
      "\n",
      " Second law efficiency is  6.0  percent\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "p1 = 1 # Air pressure at compressure inlet in bar\n",
    "t1 = 30 # Air temperature at compressure inlet in degree Celsius\n",
    "p2 = 3.5 # Air pressure at compressure exit in bar\n",
    "t2 = 141 # Air temperature at compressure exit in degree Celsius\n",
    "v = 90 # Air velocity at compressure exit in m/s\n",
    "cp = 1.0035 # Specific heat capacity of air in kJ/kg\n",
    "y = 1.4 # Heat capacity ratio\n",
    "R = 0.287 # Gas constant\n",
    "print \"\\n Example 8.20\\n\"\n",
    "T2s = (t1+273)*(p2/p1)**((y-1)/y)\n",
    "if T2s>(t2+273): \n",
    "    print \"\\n Part A:\"\n",
    "    print \"\\n There is heat loss to surrounding.\"\n",
    "n =(1/(1-((math.log((t2+273)/(t1+273)))/(math.log(p2/p1)))))\n",
    "print \"\\n\\n Part B:\"\n",
    "print \"\\n The polytropic index is \",n\n",
    "Wa = cp*(t1-t2)-(v**2)/2000 # Actual work \n",
    "Wt = -R*(t1+273)*math.log(p2/p1) - (v**2)/2000 # Isothermal work\n",
    "nt =Wt/Wa # Isothermal efficency\n",
    "print \"\\n\\n Part C:\"\n",
    "print \"\\n Isothermal efficiency is \",nt*100 ,\" percent \"\n",
    "df = cp*(t1-t2) + (t1+273)*(R*math.log(p2/p1) - cp*math.log((t2+273)/(t1+273))) -(v**2)/2000\n",
    "Wm = df # Minimum work input\n",
    "I = Wm-Wa # Irreversibility\n",
    "\n",
    "print \"\\n\\n Part D:\"\n",
    "print \"\\n The minimum work input is \",Wm,\" kJ/kg, and irreversibility is \",I ,\" kJ/kg\"\n",
    "# The answers given in the book contain round off error\n",
    "\n",
    "neta = Wm/Wa\n",
    "print \"\\n\\n Part E:\"\n",
    "print \"\\n Second law efficiency is \",math. ceil(neta*100) ,\" percent\"\n",
    "\n"
   ]
  }
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