{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#Chapter 1: Definitions and Basic Relations"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###Example 1.1, Internal Energy and Enthalpy, Page No. 38"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Enthalpy = 301.500000 kJ/kg\n",
      " Internal Energy = 215.400000 kJ/kg\n",
      "\n",
      " Diameter = 46.824337 cm\n"
     ]
    }
   ],
   "source": [
    "from math import pi;\n",
    "from math import sqrt;\n",
    "#variable declaration\n",
    "R=0.287;      #in kJ.kg K\n",
    "c_p=1.005;      #in kJ.kg K\n",
    "m=3;         #in kg/s\n",
    "T=300;       #in K\n",
    "p=1.5;        #in bar\n",
    "c=10;         #in m/s\n",
    "p=p*10**5;    #converts bar into Pa\n",
    "\n",
    "#calculation\n",
    "c_v=c_p-R;\n",
    "h=c_p*T;\n",
    "u=c_v*T;\n",
    "rho=p/(R*T*1000);\n",
    "D=sqrt((4*m)/(pi*c*rho));\n",
    "D=D*100;   #converts m into cm\n",
    "\n",
    "#result\n",
    "print('\\n Enthalpy = %f kJ/kg\\n Internal Energy = %f kJ/kg')%(h,u);\n",
    "print '\\n Diameter = %f cm' %(D);\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###Example 1.2, Flow and Non-Flow Work, Page No. 38"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Final Temperature = 382.397541 K\n",
      " Enthalpy Drop = 88.473301 kJ/kg\n",
      " Change in Internal Energy = 71.349436 kJ/kg\n"
     ]
    }
   ],
   "source": [
    "#variable declaration\n",
    "R=0.189;       #in kJ/kg K\n",
    "gamma_1=1.24;   #no unit\n",
    "T1=473;        #in K\n",
    "p1=3.0;          #in bar\n",
    "p2=1.0;          #in bar\n",
    "\n",
    "#calculation\n",
    "c_p=(gamma_1*R)/(gamma_1-1);\n",
    "c_v=c_p/gamma_1;\n",
    "ratio=(p2/p1)**((gamma_1-1)/gamma_1);\n",
    "T2=ratio*T1;\n",
    "h=c_p*(T1-T2);\n",
    "u=c_v*(T1-T2);\n",
    "\n",
    "#result\n",
    "print('\\n Final Temperature = %f K\\n Enthalpy Drop = %f kJ/kg\\n Change in Internal Energy = %f kJ/kg')%(T2,h,u);\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###Example 1.3, Change of Entropy in a Polytropic Process, Page No. 39"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "Change in Entropy = 0.165932\n",
      "\n",
      "\n",
      "Note : There are computational problems in the book of this example\n"
     ]
    }
   ],
   "source": [
    "#variable declaration\n",
    "from math import log;\n",
    "gamma_1=1.3;      #no unit\n",
    "T1=650;       #in K\n",
    "n=1.2;       #no unit\n",
    "p1=10.0;      #in bar\n",
    "p2=3.0;        #in bar\n",
    "c_p=2.15;      #in kJ/kg K\n",
    "\n",
    "#cslculation\n",
    "c_v=c_p/gamma_1;\n",
    "ratio_p=p2/p1;\n",
    "ratio_v=(1/ratio_p)**(1/n);\n",
    "s=c_v*log(ratio_p)+c_p*log(ratio_v);\n",
    "\n",
    "#result\n",
    "print('\\nChange in Entropy = %f')%(s);\n",
    "print('\\n\\nNote : There are computational problems in the book of this example')\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###Example 1.4, Fanning's Coefficient, Page No. 39"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "Pressure at the exit of duct = 2.925336 bar\n",
      "\n",
      "\n",
      "Note : There are computational problems in the book of this example\n"
     ]
    }
   ],
   "source": [
    "#variable declaration\n",
    "L=100;     #in m\n",
    "R=287;      #in kJ/kg K\n",
    "D=0.5;       #in m\n",
    "T=315;       #in K\n",
    "p=3.0;         #in bar\n",
    "c=15;        #in m/s\n",
    "f=0.025;      #no unit\n",
    "\n",
    "#calculation\n",
    "rho=p/(R*T);\n",
    "delta_p=4*f*L*rho*c**2/(2*D)\n",
    "p2=p-delta_p;\n",
    "\n",
    "#result\n",
    "print('\\nPressure at the exit of duct = %f bar')%(p2);\n",
    "print('\\n\\nNote : There are computational problems in the book of this example')\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###Example 1.5, Adiabatic Bulk Modulus of a Gas, Page No. 40"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Bulk Modulus of Elasticity = 31295.652174 bar\n",
      "\n",
      " More Accurate Value of Bulk Modulus of Elasticity = 29459.521175 bar\n"
     ]
    }
   ],
   "source": [
    "#variable declaration\n",
    "p1=1;         #in bar\n",
    "p2=3600;      #in bar\n",
    "v1=1;         #in m^3\n",
    "v2=0.885       #in m^3\n",
    "\n",
    "#calculation & result\n",
    "K_t=-v1*(p2-p1)/(v2-v1);\n",
    "print('\\n Bulk Modulus of Elasticity = %f bar')%(K_t);\n",
    "K_t=(p2-p1)/log(v1/v2);\n",
    "print('\\n More Accurate Value of Bulk Modulus of Elasticity = %f bar')%(K_t)\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###Example 1.6, Conditions at altitudes, Page No. 41"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "At Z=10000m\n",
      "Temperature = 223.200000 K\n",
      "Pressure = 0.264259 bar\n",
      "Density = 0.412528 kg/m**3\n",
      "Viscosity = 0.0000148 kg/ms\n",
      "\n",
      "At Z=15000m\n",
      "Temperature = 216.500000 K\n",
      "Pressure = 0.120714 bar\n",
      "Density = 0.194276 kg/m**3\n",
      "Viscosity = 0.0000142 kg/ms\n"
     ]
    }
   ],
   "source": [
    "#variable declaration\n",
    "from math import exp;\n",
    "p0=1.0133;       #in bar\n",
    "#p0=p0*10**5;      #conversion to Pa\n",
    "T0=288.2;         # in K\n",
    "Tt=216.5;         # in K\n",
    "u0=1.79*10**-5;     #in kg/ms\n",
    "ut=1.42*10**-5;     #in kg/ms\n",
    "pt=0.227;          #in bar\n",
    "Z1=10000;          #in m\n",
    "Z2=15000;          #in m\n",
    "Zt=11000;          #in m\n",
    "R=287;           #in J/kg K\n",
    "a1=6.5/1000;         #in deg C/m\n",
    "g=9.81;           #in m/s**2\n",
    "\n",
    "#calculation\n",
    "rho0=p0/(R*T0);\n",
    "T=T0-a1*Z1;\n",
    "p=p0*(T/T0)**(g/(a1*R));\n",
    "rho=p*10**5/(R*T);\n",
    "u=u0*(T/T0)**0.75;\n",
    "p1=pt*exp(-g*(Z2-Zt)/(R*Tt));\n",
    "rho1=p1*10**5/(R*Tt);\n",
    "\n",
    "#result\n",
    "print('\\nAt Z=10000m\\nTemperature = %f K\\nPressure = %f bar\\nDensity = %f kg/m**3\\nViscosity = %.7f kg/ms\\n\\nAt Z=15000m\\nTemperature = %f K\\nPressure = %f bar\\nDensity = %f kg/m**3\\nViscosity = %.7f kg/ms')%(T,p,rho,u,Tt,p1,rho1,ut);\n",
    "\n"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 2",
   "language": "python",
   "name": "python2"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 2
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython2",
   "version": "2.7.10"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}