{
 "metadata": {
  "name": "Chapter 2"
 }, 
 "nbformat": 2, 
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown", 
     "source": [
      "#Chapter 2:- Energy and the First Law Of Thermodynamics"
     ]
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading5> Example 2.1 Page no-39"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "p1 = 3*(10**5)                                # initial pressure of gas in pascal", 
      "v1 = 0.1                                      # initial volumme of gas in meter^3", 
      "v2 = 0.2                                      # final volume of gas in meter^3", 
      "", 
      "# calculations", 
      "from scipy import integrate", 
      "import math", 
      "# Part (a) i.e. n=1.5", 
      "#constant = p1*(v1**n)                         # p*(v^n) = constant", 
      "constant1 = p1*(v1**1.5) ", 
      "constant2 = p1*(v1**1) ", 
      "constant3 = p1*(v1**0) ", 
      "# function p                     ", 
      "p1 = lambda v: constant1/(v**1.5)                   # expressing pressure as function of volume        ", 
      "p2 = lambda v: constant2/(v**1)", 
      "p3 = lambda v: constant3/(v**0)", 
      "# function [work1] ", 
      "work1 = integrate.quad(p1,v1,v2)                   # integrating pdv from initial to final volume ", 
      "w1 = work1[0]/1000                             # divided by 1000 to convert to KJ", 
      "print 'The work done for n=1.5 in KJ is',round(w1,2)", 
      "", 
      "#part(b) i.e. n = 1", 
      "work2 = integrate.quad(p2,v1,v2)", 
      "w2 = work2[0]/1000", 
      "print 'The work done for n=1 in KJ is',round(w2,2)", 
      "", 
      "#part(c) i.e. n=0", 
      "work3 = integrate.quad(p3,v1,v2)", 
      "w3 = work3[0]/1000", 
      "print 'The work done for n=0 in KJ is',round(w3,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "The work done for n=1.5 in KJ is 17.57", 
        "The work done for n=1 in KJ is 20.79", 
        "The work done for n=0 in KJ is 30.0"
       ]
      }
     ], 
     "prompt_number": 1
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 2.2 Page no-50"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "p1 = 3*(10**5)                                  # initial pressure in pascal", 
      "v1 = 0.1                                        # initial volume in m3", 
      "v2 = 0.2                                        # initial volume in m3", 
      "m = 4.0                                         # mass of the gas in kg", 
      "deltau = -4.6                                   # change in specific internal energy in KJ/Kg", 
      "", 
      "# Calculations", 
      "from scipy import integrate", 
      "import math", 
      "constant = p1*(v1**1.5)                           # p*(v^n) = constant", 
      "", 
      "p = lambda v: constant/(v**1.5)                                # expressing pressure as function of volume        ", 
      "", 
      "work = integrate.quad(p,v1,v2)                         # integrating pdv from initial to final volume ", 
      "w=work[0]/1000                              # divided by 1000 to convert to KJ", 
      "", 
      "deltaU = m*deltau                           # change in internal energy in KJ", 
      "Q = deltaU + w                                # neglecting kinetic and potential energy changes", 
      "", 
      "# Result", 
      "print 'net heat transfer for the process in KJ',round(Q,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "net heat transfer for the process in KJ -0.83"
       ]
      }
     ], 
     "prompt_number": 2
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 2.3 Page no-52"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "patm = 10**5                               # atmospheric pressure in pascal.", 
      "mp = 45.0                                  # mass of piston in Kg", 
      "A = 0.09                                   # face area of piston in m2", 
      "deltaV = 0.045                             # increment of the volume of air in m3", 
      "m = 0.27                                   # mass of air in kg", 
      "deltau = 42.0                              # specific internal energy increase of air in kJ/kg", 
      "g = 9.81                                   # local acceleration of gravity", 
      "", 
      "", 
      "# Part (a) i.e. air is system", 
      "# Calculations", 
      "p = (mp*g)/A + patm                         # constant pressure of air obtained from equilibrium of piston", 
      "w = (p*deltaV)/1000                         # work done in KJ", 
      "deltaU = m*deltau                           # internal energy change of air in KJ", 
      "Q = w + deltaU                              # applying first with air as system", 
      "# Result", 
      "print 'The answer given in book is incorrect.They have miscalculated deltaU.The correct heat transfer from resistor to air in KJ for air alone as system is: '", 
      "print round(Q,2)", 
      "", 
      "# The answer given in book is incorrect. deltaU is incorrect in book. ", 
      "", 
      "# Part(b) i.e.  (air+piston) is system", 
      "# Calculations", 
      "wd = (patm*deltaV)/1000                      # work done in KJ", 
      "deltaz = (deltaV)/A                          # change in elevation of piston", 
      "deltaPE = (mp*g*deltaz)/1000                 # change in potential energy of piston in KJ", 
      "Qt = wd + deltaPE + deltaU                   # applying first law with air plus piston as system ", 
      "# Result", 
      "print 'The answer given in book is incorrect.They have miscalculated deltaU.The correct heat transfer from resistor to air in KJ for air + piston as system is:'", 
      "print round(Qt,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "The answer given in book is incorrect.They have miscalculated deltaU.The correct heat transfer from resistor to air in KJ for air alone as system is: ", 
        "16.06", 
        "The answer given in book is incorrect.They have miscalculated deltaU.The correct heat transfer from resistor to air in KJ for air + piston as system is:", 
        "16.06"
       ]
      }
     ], 
     "prompt_number": 3
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 2.4 Page no-54"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "w1dot = -60.0                # input work rate in KW", 
      "h = 0.171                    # heat transfer coefficient,unit in KW/m2 .K", 
      "A = 1.0                      # outer surface area of gearbox, unit in m2", 
      "Tb = 300.0                   # outer surface temperature in kelvin", 
      "Tf = 293.0                   # temperature of the sorrounding", 
      "", 
      "# Calculations", 
      "Qdot = -h*A*(Tb-Tf);         # rate of energy transfer by heat", 
      "wdot = Qdot;                 # steady state energy equation", 
      "w2dot = wdot-w1dot;", 
      "", 
      "# Results", 
      "print 'The heat transfer rate in KW is:\\n\\tQdot =  ',Qdot", 
      "print 'The power delivered through output shaft in KW is: = ',w2dot"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "The heat transfer rate in KW is:", 
        "\tQdot =   -1.197", 
        "The power delivered through output shaft in KW is: =  58.803"
       ]
      }
     ], 
     "prompt_number": 4
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 2.5 Page no-55"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "s=5*(10**-3)                       # measurement on a side in meter", 
      "wdot = -0.225                        # power input in watt", 
      "Tf = 293.0                           # coolant temprature in kelvin", 
      "h = 150.0                            # heat transfer coefficient in w/m2 k", 
      "A = s**2                            # surface area", 
      "", 
      "# Calculation", 
      "Tb = ((-wdot/(h*A)) + Tf - 273)     # surface temperature in degree", 
      "", 
      "# Result", 
      "print 'The surface temperature of the chip in degree celcius is:  ',Tb"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "The surface temperature of the chip in degree celcius is:   80.0"
       ]
      }
     ], 
     "prompt_number": 5
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 2.6 Page no-57"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "%matplotlib inline"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "", 
        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].", 
        "For more information, type 'help(pylab)'."
       ]
      }
     ], 
     "prompt_number": 6
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "omega = 100.0                      #motor rotation speed in rad/s", 
      "tau = 18.0                         #torque applied by shaft in N.m", 
      "Welecdot = -2.0                    #electric power input in KW", 
      "", 
      "Wshaftdot = (tau*omega)/1000      #shaft work rate in KW", 
      "Wdot = Welecdot + Wshaftdot       #net work rate in KW", 
      "", 
      "#function [Qdot]=f(t)", 
      "#Qdot = (-0.2)* [1-2**(-0.05*t)]", 
      "", 
      "", 
      "#function [Edot]=f1(t)               #function for rate of change of energy", 
      "#Edot = (-0.2)*[1-2**(-0.05*t)] - Wdot ", 
      "", 
      "#function [deltaE] =f2(t)            #function for change in energy  ", 
      "from sympy import *", 
      "x = symbols('x')             # a = 0.2 b = 0.05", 
      "f = (0.2)*(1-2**(0.05*(x))) - Wdot", 
      "f2 = integrate(f,x)", 
      "print f2         ", 
      "", 
      "Qd = []", 
      "Wd = []", 
      "dltaE = []", 
      "from numpy import linspace ", 
      "from pylab import plot, show ", 
      "t = linspace(0,120,100);", 
      "for i in range(0,100):", 
      "    Qd.append(i)", 
      "    Wd.append(i)", 
      "    dltaE.append(i)", 
      "    Qd[i] =  (-0.2*(1-2**(-0.05*(120/99)*(i-1))))                                     ", 
      "    Wd[i] = Wdot", 
      "    dltaE[i] = -4.0*2**(0.05* (120/99)*(i-1))/log(2) + 0.4* (120/99)*(i-1)", 
      "                          ", 
      "plot(t,Qd)", 
      "xlabel(\"Time (s)\")", 
      "ylabel(\"Qdot (KW)\")", 
      "show()", 
      "plot(t,Wd)", 
      "xlabel(\"Time (s)\")", 
      "ylabel(\"Wdot (KW)\")", 
      "show()", 
      "plot(t,dltaE)", 
      "xlabel(\"Time (s)\")", 
      "ylabel(\"deltaE (KJ)\")", 
      "show()", 
      "", 
      "", 
      "", 
      "", 
      "     "
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-4.0*2**(0.05*x)/log(2) + 0.4*x"
       ]
      }, 
      {
       "output_type": "display_data", 
       "png": 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      }, 
      {
       "output_type": "display_data", 
       "png": 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      }
     ], 
     "prompt_number": 7
    }
   ]
  }
 ]
}