path: root/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter13.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 13: Gas power cycle"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.1:pg-554"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.1\n",
      "\n",
      "\n",
      " Cycle efficiency is  56.4724718352  percent\n",
      "\n",
      " Maximum temperature in the cycle is  3632.38927303  K\n",
      "\n",
      " Maximum pressure in the cycle is  9.43477733254  MPa\n",
      "\n",
      " Mean effective pressure is  1.53325865881  MPa\n"
     ]
    }
   ],
   "source": [
    "T1 = 35 # Air inlet temperature in degree Celsius\n",
    "P1 = 0.1 # Air inlet pressure in MPa\n",
    "Q1 = 2100 # Heat supply in kJ/kg\n",
    "R = 0.287 # gas constant\n",
    "rk = 8 # Compression ratio\n",
    "g = 1.4 # Heat capacity ratio\n",
    "n_cycle = 1-(1/rk**(g-1)) # cycle efficiency \n",
    "v1 = (R*(T1+273))/(P1*1e3) # Initial volume\n",
    "v2 = v1/8 # Volume after compression\n",
    "T2 = (T1+273)*(v1/v2)**(g-1) # Temperature after compression\n",
    "cv = 0.718 # Constant volume heat capacity in kJ/kg\n",
    "T3 = Q1/cv + T2 # Temperature at after heat addition\n",
    "P21 = (v1/v2)**g # Pressure ratio\n",
    "P2 = P21*P1 # Pressure after compression\n",
    "P3 = P2*(T3/T2) # Pressure after heat addition\n",
    "Wnet = Q1*n_cycle # Net work output\n",
    "Pm = Wnet/(v1-v2) # Mean pressure\n",
    "print \"\\n Example 13.1\\n\"\n",
    "print \"\\n Cycle efficiency is \",n_cycle*100 ,\" percent\"\n",
    "print \"\\n Maximum temperature in the cycle is \",T3 ,\" K\"\n",
    "print \"\\n Maximum pressure in the cycle is \",P3 ,\" MPa\"\n",
    "print \"\\n Mean effective pressure is \",Pm/1e3 ,\" MPa\"\n",
    "#The answers vary due to round off error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.2:pg-555"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.2\n",
      "\n",
      "\n",
      " Air standard efficiency is  59.8676909231  percent\n"
     ]
    }
   ],
   "source": [
    "\n",
    "rk = 14.0 # Compression ratio\n",
    "k = 6.0 # cutoff percentage ratio\n",
    "rc = k/100*(rk-1)+1\n",
    "g = 1.4 # Heat capacity ratio\n",
    "n_diesel = 1.0-((1.0/g))*(1.0/rk**(g-1))*((rc**(g-1))/(rc-1)) # Cycle efficiency\n",
    "print \"\\n Example 13.2\\n\"\n",
    "print \"\\n Air standard efficiency is \",n_diesel*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.3:pg-556"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.3\n",
      "\n",
      "\n",
      " Cut-off ratio is  2.00789702047\n",
      "\n",
      " Heat supplied per kg of air is  884.346993978  kJ/kg\n",
      "\n",
      " Cycle efficiency is  61.3340410825  percent\n",
      "\n",
      " Mean effective pressure is  699.968703831  kPa\n"
     ]
    }
   ],
   "source": [
    "rk = 16 # Compression ratio\n",
    "T1 = 15 # Air inlet temperature in degree Celsius\n",
    "P1 = 0.1 # Air inlet pressure in MPa\n",
    "T3 = 1480 # Highest temperature in cycle in degree Celsius\n",
    "g = 1.4 # Heat capacity ratio\n",
    "R = 0.287 # Gas constant\n",
    "T2 = (T1+273)*(rk**(g-1)) # Temperature after compression\n",
    "rc =  (T3+273)/T2 # cut off ratio\n",
    "cp = 1.005 # Constant pressure heat constant\n",
    "cv = 0.718 # Constant volume heat constant\n",
    "Q1 = cp*(T3+273-T2) # Heat addition\n",
    "T4 = (T3+273)*((rc/rk)**(g-1)) # Temperature after heat addition\n",
    "Q2 = cv*(T4-T1-273) # Heat rejection\n",
    "n = 1-(Q2/Q1) # cycle efficiency\n",
    "n_ = 1-((1/g))*(1/rk**(g-1))*((rc**(g-1))/(rc-1)) # cycle efficiency from another formula\n",
    "Wnet = Q1*n # Net work \n",
    "v1 = (R*(T1+273))/(P1*1e3) # Volume before compression\n",
    "v2 = v1/rk # Volume after compression\n",
    "Pm = Wnet/(v1-v2) # Mean pressure\n",
    "print \"\\n Example 13.3\\n\"\n",
    "print \"\\n Cut-off ratio is \",rc\n",
    "print \"\\n Heat supplied per kg of air is \",Q1 ,\" kJ/kg\"\n",
    "print \"\\n Cycle efficiency is \",n*100 ,\" percent\"\n",
    "print \"\\n Mean effective pressure is \",Pm ,\" kPa\"\n",
    "#The answers vary due to round off error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.4:pg-558"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.4\n",
      "\n",
      "\n",
      " Efficiency of the cycle is  66.3143793932  percent\n",
      "\n",
      " Mean effective pressure is  4.45799460092  bar\n"
     ]
    }
   ],
   "source": [
    "T1 = 50.0 # Temperature before compression stroke in degree Celsius\n",
    "rk = 16.0 # Compression ratio\n",
    "g = 1.4 # Heat capacity ratio\n",
    "P3 = 70.0 # Maximum cycle pressure in bar\n",
    "cv = 0.718 # Constant volume heat addition capacity\n",
    "cp = 1.005 # Constant pressure heat addition capacity\n",
    "R = 0.287 # Gas constant\n",
    "T2 = (T1+273)*((rk**(g-1))) #Temperature after compression stroke \n",
    "P1 = 1.0 # Pressure before compression in bar\n",
    "P2 = P1*(rk)**g # Pressure after compression\n",
    "T3 = T2*(P3/P2) # Temperature after constant volume heat addition\n",
    "Q23 = cv*(T3-T2) # Constant volume heat added\n",
    "T4 = (Q23/cp)+T3 # Temperature after constant pressure heat addition\n",
    "v43 = T4/T3 # cut off ratio \n",
    "v54 = rk/v43 # Expansion ratio\n",
    "T5 = T4*(1/v54)**(g-1) # Temperature after expansion\n",
    "P5 = P1*(T5/(T1+273)) # Pressure after expansion\n",
    "Q1 = cv*(T3-T2)+cp*(T4-T3) # Total heat added\n",
    "Q2 = cv*(T5-T1-273) # Heat rejected\n",
    "n_cycle = 1-(Q2/Q1) # Cycle efficiency\n",
    "v1 = (R*(T1+273))/(P1*1e2) # Volume before compression \n",
    "v2 = (1/16)*v1 # Swept volume\n",
    "Wnet = Q1*n_cycle # Net work done\n",
    "Pm = Wnet/(v1-v2) # Mean pressure\n",
    "print \"\\n Example 13.4\\n\"\n",
    "print \"\\n Efficiency of the cycle is \",n_cycle*100 ,\" percent\"\n",
    "print \"\\n Mean effective pressure is \",Pm/100 ,\" bar\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.5:pg-559"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.5\n",
      "\n",
      "\n",
      " The percentage increase in cycle efficiency \n",
      " due to regeneration is  41.4076056717  percent\n"
     ]
    }
   ],
   "source": [
    "P1 = 0.1 # Air pressure at turbine inlet in MPa\n",
    "T1 = 30 # Air temperature at turbine inlet in degree Celsius\n",
    "T3 = 900 # Maximum cycle temperature at turbine inlet in degree Celsius\n",
    "rp = 6 # Pressure ratio\n",
    "nt = 0.8 # Turbine efficiency\n",
    "nc = 0.8# Compressor efficiency\n",
    "g = 1.4 # Heat capacity ratio\n",
    "cv = 0.718 # Constant volume heat capacity\n",
    "cp = 1.005 # Constant pressure heat capacity\n",
    "R = 0.287 # Gas constant\n",
    "T2s = (T1+273)*(rp)**((g-1)/g)\n",
    "T4s = (T3+273)/((rp)**((g-1)/g))\n",
    "T21 = (T2s-T1-273)/nc  # Temperature raise due to compression\n",
    "T34 = nt*(T3+273-T4s) # Temperature drop due to expansion\n",
    "Wt = cp*T34 # Turbine work\n",
    "Wc = cp*T21 # Compressor work\n",
    "T2 = T21+T1+273 # Temperature after compression\n",
    "Q1 = cp*(T3+273-T2) # Heat added\n",
    "n = (Wt-Wc)/Q1 # First law efficiency\n",
    "T4 = T3+273-T34 # Temperature after expansion\n",
    "T6 = 0.75*(T4-T2) + T2 # Regeneration temperature \n",
    "Q1_ = cp*(T3+273-T6)# Heat added\n",
    "n_ = (Wt-Wc)/Q1_ #cycle efficiency\n",
    "I = (n_-n)/n # Fractional increase in cycle efficiency\n",
    "print \"\\n Example 13.5\\n\"\n",
    "print \"\\n The percentage increase in cycle efficiency \\n due to regeneration is \",I*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.6:pg-560"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.6\n",
      "\n",
      "\n",
      " Maximum work done per kg of air is  239.466740619  kJ/kg\n",
      "\n",
      " Cycle efficiency is  47.1237354986  percent\n",
      "\n",
      " Ratio of Brayton and Carnot efficiency is  0.654123779948\n"
     ]
    }
   ],
   "source": [
    "\n",
    "cp = 1.005 # Constant pressure heat capacity\n",
    "Tmax = 1073.0 # Maximum cycle temperature in K\n",
    "Tmin = 300.0# Minimum cycle temperature in K\n",
    "Wnet_max = cp*(sqrt(Tmax)-sqrt(Tmin))**2 # maximum work\n",
    "n_cycle = 1.0-sqrt(Tmin/Tmax) # cycle efficiency\n",
    "n_carnot = 1.0-(Tmin/Tmax) # Carnot efficiency\n",
    "r = n_cycle/n_carnot # Efficiency ratio\n",
    "print \"\\n Example 13.6\\n\"\n",
    "print \"\\n Maximum work done per kg of air is \",Wnet_max ,\" kJ/kg\"\n",
    "print \"\\n Cycle efficiency is \",n_cycle*100 ,\" percent\"\n",
    "print \"\\n Ratio of Brayton and Carnot efficiency is \",r\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.7:pg-561"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.7\n",
      "\n",
      "\n",
      " The thermal efficiency of the cycle is  40.0663025288  percent\n",
      "\n",
      " Work ratio is  0.544951697902\n",
      "\n",
      " Power output is  40.0663025288  MW\n",
      "\n",
      " Energy flow rate of the exhaust gas stream is  20.5297861501  MW\n"
     ]
    }
   ],
   "source": [
    "\n",
    "rp = 6 # pressure ratio\n",
    "g = 1.4 # Heat capacity ratio\n",
    "cv = 0.718 # Constant volume heat capacity\n",
    "cp = 1.005 #Constant pressure heat capacity\n",
    "R = 0.287 # Gas constant\n",
    "T1 = 300 # Minimum temperature in K\n",
    "T3 = 1100 # Maximum cycle temperature in K\n",
    "T0 = 300 # Atmospheric temperature in K\n",
    "n_cycle = 1-(1/rp**((g-1)/g)) # cycle efficiency\n",
    "T2 = (T1)*(rp**((g-1)/g)) # Temperature after compression\n",
    "T4 = (T3)/(rp**((g-1)/g)) # Temperature after expansion\n",
    "Wc = cp*(T2-T1) # Compressor work\n",
    "Wt = cp*(T3-T4) # Turbine work\n",
    "WR = (Wt-Wc)/Wt # Work ratio\n",
    "Q1 = 100 # Heat addition in MW\n",
    "PO = n_cycle*Q1 # Power output\n",
    "m_dot = (Q1*1e06)/(cp*(T3-T2)) # Mass flow rate\n",
    "R = m_dot*cp*T0*((T4/T0)-1-log(T4/T0)) # Exergy flow rate\n",
    "print \"\\n Example 13.7\\n\"\n",
    "print \"\\n The thermal efficiency of the cycle is \",n_cycle*100 ,\" percent\"\n",
    "print \"\\n Work ratio is \",WR\n",
    "print \"\\n Power output is \",PO ,\" MW\"\n",
    "print \"\\n Energy flow rate of the exhaust gas stream is \",R/1e6 ,\" MW\"\n",
    "#The answers vary due to round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.8:pg-562"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.8\n",
      "\n",
      "\n",
      " Percentage of air that may be taken from the compressor is  11.5044247788  percent\n"
     ]
    }
   ],
   "source": [
    "nc = 0.87 # Compressor efficiency \n",
    "nt = 0.9 # Turbine efficiency\n",
    "T1 = 311 # Compressor inlet temperature in K\n",
    "rp = 8 # compressor pressure ratio\n",
    "P1 = 1 # Initial pressure in atm\n",
    "T3 = 1367 # Turbine inlet temperature\n",
    "P2 = P1*rp # Final pressure \n",
    "P3 = 0.95*P2 # Actual pressure after compression\n",
    "P4 = 1 # Atmospheric pressure\n",
    "g = 1.4 # Heat capacity ratio\n",
    "cv = 0.718 # Constant volume heat capacity\n",
    "cp = 1.005 # Constant pressure heat capacity\n",
    "R = 0.287 # Gas constant\n",
    "# With no cooling\n",
    "T2s = T1*((P2/P1)**((g-1)/g)) # Ideal temperature after compression\n",
    "T2 = T1 + (T2s-T1)/0.87 # Actual temperature after compression\n",
    "T4s = T3*(P4/P3)**((g-1)/g) # Ideal temperature after expansion\n",
    "n = (((T3-T4s)*nt)-((T2s-T1)/nc))/(T3-T2) # cycle efficiency\n",
    "# With cooling\n",
    "n_cycle = n-0.05\n",
    "x = 0.13 # Fluid quality\n",
    "r = x/(x+1) # \n",
    "print \"\\n Example 13.8\\n\"\n",
    "print \"\\n Percentage of air that may be taken from the compressor is \",r*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.9:pg-563"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.9 \n",
      "\n",
      "\n",
      " Optimum specific output is  1.0\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#Given that\n",
    "nc = 0.85 # Compressor efficiency\n",
    "nt = 0.9 # Turbine efficiency\n",
    "r = 3.5 # Ratio of max and min temperature \n",
    "gama = 1.4 # Ratio of heat capacities for air\n",
    "print \"\\n Example 13.9 \\n\"\n",
    "x = (gama-1)/gama\n",
    "r_opt = ((nc*nt*r)**(2/3))**(1/x)\n",
    "print \"\\n Optimum specific output is \",r_opt\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex13.10:pg-566"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 13.10 \n",
      "\n",
      "\n",
      " The temperature of the gases at the turbine exit is  1114.47439653  K,\n",
      " The pressure of the gases at the turbine exit is  311.998817219  kN/m**2,\n",
      " The velocity of gases at the nozzle exit is  1.0  m/sec,\n",
      " The propulsive efficiency of the cycle is  -10.6673736259  percent\n"
     ]
    }
   ],
   "source": [
    "#Given that\n",
    "v = 300.0 # Aircraft velocity in m/s\n",
    "p1 = 0.35 # Pressure in bar\n",
    "t1 = -40.0 # Temperature in degree centigrade\n",
    "rp = 10.0 # The pressure ratio of compressor \n",
    "t4 = 1100.0 # Temperature of gases at turbine intlet in degree centigrade\n",
    "ma = 50.0 # Mass flow rate of air at the inlet of compressor in kg/s\n",
    "cp = 1.005 # Heat capacity of air at constant pressure in kJ/kg-K\n",
    "gama=1.4 # Ratio of heat capacities for air\n",
    "print \"\\n Example 13.10 \\n\"\n",
    "T1 = t1+273\n",
    "T4 = t4+273\n",
    "T2 = T1 + (v**2)/(2*cp)*(10**-3)\n",
    "p2 = p1*(100)*((T2/T1)**(gama/(gama-1)))\n",
    "p3 = rp*p2\n",
    "p4 =p3\n",
    "T3 = T2*((p3/p2)**((gama-1)/gama))\n",
    "T5 = T4-T3+T2\n",
    "p5 = ((T5/T4)**(gama/(gama-1)))*(p4)\n",
    "p6 = p1*100\n",
    "T6 = T5*((p6/p5)**((gama-1)/gama))\n",
    "V6 = (2*cp*(T5-T6)*1000)**(1/2)\n",
    "Wp = ma*(V6-v)*v*(10**-6)\n",
    "Q1 = ma*cp*(T4-T3)*(10**-3)\n",
    "np = Wp/Q1\n",
    "print \"\\n The temperature of the gases at the turbine exit is \",T5 ,\" K,\\n The pressure of the gases at the turbine exit is \",p5 ,\" kN/m**2,\\n The velocity of gases at the nozzle exit is \",V6 ,\" m/sec,\\n The propulsive efficiency of the cycle is \",np*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  }
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