path: root/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02.ipynb

{
 "metadata": {
  "name": "",
  "signature": "sha256:79d03e9c97c3d8ac9e7b819d4383e36696e6a2ea64bb801edb77a95675cdc73c"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 02: Heat, Work, Internal Energy, Enthalpy, and The First Law of Thermodynamics"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 2.1, Page 18"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Variable Declaration  Part a\n",
      "vi = 20.0             #Initial volume of ideal gas, L\n",
      "vf = 85.0             #final volume of ideal gas, L\n",
      "Pext = 2.5            #External Pressure against which work is done, bar\n",
      "\n",
      "#Calculations\n",
      "w = -Pext*1e5*(vf-vi)*1e-3\n",
      "\n",
      "#Results\n",
      "print 'Part a: Work done in expansion is %6.1f kJ'%(w/1000)\n",
      "\n",
      "#Variable Declaration  Part b\n",
      "ri = 1.00             #Initial diameter of bubble, cm\n",
      "rf = 3.25             #final diameter of bubble, cm\n",
      "sigm = 71.99          #Surface tension, N/m\n",
      "\n",
      "from math import pi\n",
      "#Calculations\n",
      "w = -2*sigm*4*pi*(rf**2-ri**2)*1e-4\n",
      "\n",
      "#Results\n",
      "print 'Part b: Work done in expansion of bubble is %4.2f J'%w\n",
      "\n",
      "#Variable Declaration  Part c\n",
      "i = 3.20             #Current through heating coil, A \n",
      "v = 14.5             #fVoltage applied across coil, volts\n",
      "t = 30.0             #time for which current is applied,s\n",
      "\n",
      "from math import pi\n",
      "#Calculations\n",
      "w = v*i*t\n",
      "\n",
      "#Results\n",
      "print 'Part c: Work done in paasing the cuurent through coil is %4.2f kJ'%(w/1000)\n",
      "\n",
      "#Variable Declaration  Part d\n",
      "k = 100.0            #Constant in F = -kx, N/cm \n",
      "dl = -0.15            #stretch , cm\n",
      "\n",
      "from math import pi\n",
      "#Calculations\n",
      "w = -k*(dl**2-0)/2\n",
      "\n",
      "#Results\n",
      "print 'Part d: Work done stretching th fiber is %4.2f J'%w"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Part a: Work done in expansion is  -16.2 kJ\n",
        "Part b: Work done in expansion of bubble is -1.73 J\n",
        "Part c: Work done in paasing the cuurent through coil is 1.39 kJ\n",
        "Part c: Work done stretching th fiber is -1.12 J\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 2.2, Page Number 20"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration  \n",
      "m = 100.0            #Mass of water, g \n",
      "T = 100.0            #Temperature of water, \u00b0C\n",
      "Pext = 1.0           #External Pressure on assembly, bar\n",
      "x = 10.0             #percent of water vaporised at 1 bar,-\n",
      "i = 2.00             #current through heating coil, A\n",
      "v = 12.0             #Voltage applied, v\n",
      "t = 1.0e3            #time for which current applied, s \n",
      "rhol = 997           #Density of liquid, kg/m3\n",
      "rhog = 0.59          #Density of vapor, kg/m3\n",
      "\n",
      "#Calculations\n",
      "q = i*v*t\n",
      "vi = m/(rhol*100)*1e-3\n",
      "vf = m*(100-x)*1e-3/(rhol*100) + m*x*1e-3/(rhog*100)\n",
      "w = -Pext*(vf-vi)*1e5\n",
      "#Results\n",
      "print 'Heat added to the water %4.2f kJ'%(q/1000)\n",
      "print 'Work done in vaporizing liquid is %4.2f J'%w"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat added to the water 24.00 kJ\n",
        "Work done in vaporizing liquid is -1703.84 J\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 2.3, Page Number 22"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration  Part d\n",
      "m = 1.5              #mass of water in surrounding, kg \n",
      "dT = 14.2            #Change in temperature of water, \u00b0C or K\n",
      "cp = 4.18            #Specific heat of water at constant pressure, J/(g.K)\n",
      "\n",
      "#Calculations\n",
      "qp = m*cp*dT\n",
      "\n",
      "#Results\n",
      "print 'Heat removed by water at constant pressure %4.2f kJ'%qp"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat removed by water at constant pressure 89.03 kJ\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 2.4, Page Number 28"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "#Variable declaration\n",
      "n = 2.0          #moles of ideal gas\n",
      "R = 8.314        #Ideal gas constant, bar.L/(mol.K)\n",
      "#For reverssible Isothermal expansion \n",
      "Pi1 = 25.0       #Initial Pressure of ideal gas, bar\n",
      "Vi1 = 4.50       #Initial volume of ideal gas, L\n",
      "Pf1 = 4.50       #Fianl Pressure of ideal gas, bar\n",
      "Pext = 4.50      #External pressure, bar \n",
      "Pint = 11.0      #Intermediate pressure, bar\n",
      "\n",
      "#Calcualtions reverssible Isothermal expansion \n",
      "T1 = Pi1*Vi1/(n*R)\n",
      "Vf1 = n*R*T1/Pf1\n",
      "w = -n*R*T1*log(Vf1/Vi1)\n",
      "\n",
      "#Results\n",
      "print 'For reverssible Isothermal expansion'\n",
      "print 'Work done = %4.2e J'%w\n",
      "\n",
      "#Calcualtions Single step irreverssible expansion \n",
      "\n",
      "w = -Pext*1e5*(Vf1-Vi1)*1e-3\n",
      "\n",
      "#Results\n",
      "print 'For Single step reverssible expansion'\n",
      "print 'Work done = %4.2e J'%w\n",
      "\n",
      "#Calcualtions Two step irreverssible expansion \n",
      "Vint = n*R*T1/(Pint)\n",
      "w = -Pint*1e5*(Vint-Vi1)*1e-3 - Pf1*1e5*(Vf1-Vint)*1e-3\n",
      "\n",
      "#Results\n",
      "print 'For Two step reverssible expansion'\n",
      "print 'Work done = %4.2e J'%w\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "For reverssible Isothermal expansion\n",
        "Work done = -1.93e+02 J\n",
        "For Single step reverssible expansion\n",
        "Work done = -9.22e+03 J\n",
        "For Twi step reverssible expansion\n",
        "Work done = -1.29e+04 J\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 2.5, Page Number 32"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "\n",
      "#Variable declaration\n",
      "n = 2.5          #moles of ideal gas\n",
      "R = 0.08314      #Ideal gas constant, bar.L/(mol.K)\n",
      "cvm = 20.79      #Heat Capacity at constant volume, J/(mol.K)\n",
      "\n",
      "p1 = 16.6       #Pressure at point 1, bar\n",
      "v1 = 1.00       #Volume at point 1, L\n",
      "p2 = 16.6       #Pressure at point 2, bar\n",
      "v2 = 25.0       #Volume at point 2, L \n",
      "v3 = 25.0       #Volume at point 3, L\n",
      "\n",
      "#Calculations\n",
      "T1 = p1*v1/(n*R)\n",
      "T2 = p2*v2/(n*R)\n",
      "T3 = T1         #from problem statement\n",
      "        #for path 1-2\n",
      "DU12 = n*cvm*(T2-T1)\n",
      "w12 = -p1*1e5*(v2-v1)*1e-3\n",
      "q12 = DU12 - w12\n",
      "DH12 = DU12 + n*R*(T2-T1)*1e2\n",
      "\n",
      "        #for path 2-3\n",
      "w23 = 0.0\n",
      "DU23 = q23 = n*cvm*(T3-T2)\n",
      "DH23 = -DH12\n",
      "\n",
      "\n",
      "        #for path 3-1\n",
      "DU31 = 0.0       #Isothemal process\n",
      "DH31 = 0.0\n",
      "w31 = -n*R*1e2*T1*log(v1/v3)\n",
      "q31 = -w31\n",
      "\n",
      "DU = DU12+DU23+DU31\n",
      "w = w12+w23+w31\n",
      "q = q12+q23+q31\n",
      "DH = DH12+DH23+DH31\n",
      "\n",
      "#Results\n",
      "print 'For Path       q         w           DU            DH         '\n",
      "print '1-2       %7.2f  %7.2f    %7.2f     %7.2f'%(q12,w12,DU12,DH12)\n",
      "print '2-3       %7.2f    %7.2f   %7.2f    %7.2f'%(q23,w23,DU23,DH23)\n",
      "print '3-1        %7.2f    %7.2f     %7.2f       %7.2f'%(q31,w31,DU31,DH31)\n",
      "print 'Overall    %7.2f  %7.2f     %7.2f       %7.2f'%(q,w,DU,DH)\n",
      "print 'all values are in J'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "For Path       q         w           DU            DH         \n",
        "1-2       139463.96  -39840.00    99623.96     139463.96\n",
        "2-3       -99623.96       0.00   -99623.96    -139463.96\n",
        "3-1        -5343.33    5343.33        0.00          0.00\n",
        "Overall    34496.67  -34496.67        0.00          0.00\n",
        "all values are in J\n"
       ]
      }
     ],
     "prompt_number": 47
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 2.6, Page Number 34"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "n = 2.5          #moles of ideal gas\n",
      "R = 8.314        #Ideal gas constant, J/(mol.K)\n",
      "cvm = 12.47      #Heat Capacity at constant volume, J/(mol.K)\n",
      "\n",
      "pext = 1.00      #External Pressure, bar\n",
      "Ti = 325.        #Initial Temeprature, K\n",
      "pi = 2.50        #Initial Pressure, bar\n",
      "pf = 1.25        #Final pressure, bar \n",
      "\n",
      "#Calculations  Adiabatic process q = 0; DU = w\n",
      "q = 0.0     \n",
      "Tf = Ti*(cvm + R*pext/pi)/(cvm + R*pext/pf )\n",
      "DU = w = n*cvm*(Tf-Ti)\n",
      "DH = DU + n*R*(Tf-Ti)\n",
      "\n",
      "#Results\n",
      "print 'The final temperature at end of adiabatic procees is %4.1f K'%Tf\n",
      "print 'The enthalpy change of adiabatic procees is %4.1f J'%DH\n",
      "print 'The Internal energy change of  adiabatic procees is %4.1f J'%DU\n",
      "print 'The work done in expansion of adiabatic procees is %4.1f J'%w\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The final temperature at end of adiabatic procees is 268.5 K\n",
        "The enthalpy change of adiabatic procees is -2937.0 J\n",
        "The Internal energy change of  adiabatic procees is -1762.2 J\n",
        "The work done in expansion of adiabatic procees is -1762.2 J\n"
       ]
      }
     ],
     "prompt_number": 51
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 2.7, Page Number 35"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration  Part d\n",
      "h1 = 1000.0          #initial Altitude of cloud, m \n",
      "hf = 3500.0          #Final Altitude of cloud, m \n",
      "p1 = 0.802           #Pressure at h1, atm \n",
      "pf = 0.602           #Pressure at hf, atm\n",
      "T1 = 288.0           #Initial temperature of cloud, K\n",
      "cp = 28.86           #Specific heat of air, J/mol.K\n",
      "R = 8.314            #Gas constant, J/mol.K\n",
      "\n",
      "from math import log, exp\n",
      "\n",
      "#Calculations\n",
      "Tf = exp(-(cp/(cp-R)-1)/(cp/(cp-R))*log(p1/pf))*T1\n",
      "#Results\n",
      "print 'Final temperature of cloud %4.1f K'%Tf\n",
      "if Tf < 273:\n",
      "    print 'You can expect cloud'\n",
      "else:\n",
      "    print 'You can not expect cloud'\n",
      "    "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Final temperature of cloud 265.2 K\n",
        "You can expect cloud\n"
       ]
      }
     ],
     "prompt_number": 34
    }
   ],
   "metadata": {}
  }
 ]
}