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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 2 : Basic Concepts Of Thermodynamicsm"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.1 Page no : 41"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Convert the following readings of pressure to kPa assuming that barometer reads 760 mm of Hg.\n",
      "\n",
      "# Variables\n",
      "rho_Hg = 13596.; \t\t\t#kg/m**3\n",
      "g = 9.806; \t\t\t#m/s**2\n",
      "h = 0.76; \t\t\t#m\n",
      "\n",
      "# Calculations and Results\n",
      "P = rho_Hg*g*h/1000; \t\t\t#kPa\n",
      "\n",
      "\n",
      "h1 = 0.80; \t\t\t#m\n",
      "P1 = h1/h*P;\n",
      "print \"(i) Pressure of 80 cm of Hg %.3f kPa\"%P1\n",
      "\n",
      "\n",
      "print (\"(ii) 30 cm Hg vacuum\")\n",
      "H2 = 0.30; \t\t\t#cm Hg vacuum\n",
      "h2 = h-H2; \t\t\t#cm of Hg absolute\n",
      "\n",
      "P2 = h2/h*P;\n",
      "print \"Pressure due to 46 cm of Hg %.3f kPa\"%P2\n",
      "\n",
      "rho_H2O = 1000; \t\t\t#kg/m**3\n",
      "h3 = 1.35; \t\t\t#m\n",
      "P3 = rho_H2O*g*h3/1000;\n",
      "print \"(iii) Pressure due to 1.35 m H2O gauge %.3f kPa\"%(P3)\n",
      "\n",
      "\n",
      "P4 = 4.2*10**2;\n",
      "print \"(iv) 4.2 bar %.3f kPa\"%(P4),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) Pressure of 80 cm of Hg 106.658 kPa\n",
        "(ii) 30 cm Hg vacuum\n",
        "Pressure due to 46 cm of Hg 61.328 kPa\n",
        "(iii) Pressure due to 1.35 m H2O gauge 13.238 kPa\n",
        "(iv) 4.2 bar 420.000 kPa\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.2 page no : 42"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Find the pressure on the piston.\n",
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "d = 0.1; \t\t\t#m\n",
      "F = 1000.; \t\t\t#N\n",
      "\n",
      "# Calculations\n",
      "A = math.pi/4*d**2; \t\t\t#m**2\n",
      "P = F/A/10**3;\n",
      "\n",
      "# Results\n",
      "print \"Pressure on the piston = %.3f kN/m**2\"%(P),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pressure on the piston = 127.324 kN/m**2\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.3 page no : 42"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find the gauge pressure \n",
      "\n",
      "# Variables\n",
      "SG = 0.9;\n",
      "h = 1.2; \t\t\t#m\n",
      "g = 9.81; \t\t\t#m/s**2\n",
      "rho_w = 1000.; \t\t\t#kg/m**3\n",
      "\n",
      "# Calculations\n",
      "rho = SG*rho_w; \t\t\t#kg/m**3\n",
      "P = rho*g*h/10**3;\n",
      "\n",
      "# Results\n",
      "print \"Gauge pressure P = %.3f kN/m**2\"%(P)\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Gauge pressure P = 10.595 kN/m**2\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.4 page no : 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find the absolute pressure in the condenser in Pa. \n",
      "\n",
      "# Variables\n",
      "Vacuum_recorded = 740.; \t\t\t#mm of Hg\n",
      "Barometric_reading = 760.; \t\t\t#mm of Hg\n",
      "\n",
      "# Calculations\n",
      "Absolute_pressure = (Barometric_reading-Vacuum_recorded)*133.4;\n",
      "\n",
      "# Results\n",
      "print \"Absolute pressure in the condenser = %.3f Pa\"%(Absolute_pressure),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Absolute pressure in the condenser = 2668.000 Pa\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.5 page no : 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Determine :\n",
      "(i) The absolute pressure of the gas in the vessel in bar.\n",
      "(ii) Specific volume and density of the gas.\n",
      "'''\n",
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "d = 0.5; \t\t\t#m\n",
      "h = 0.75; \t\t\t#m\n",
      "m = 4.; \t\t\t#kg\n",
      "Manometer_reading = 620.; \t\t\t#mm of Hg above atmosphere\n",
      "Barometer_reading = 760.; \t\t\t#mm of Hg\n",
      "V = math.pi/4*d**2*h; \t\t\t#m**3\n",
      "print (\"(i) Total pressure in the vessel\")\n",
      "\n",
      "# Calculations and Results\n",
      "P = (Barometer_reading+Manometer_reading)*133.4/10**5; \t\t\t#bar\n",
      "print \"P = %.3f bar\"%(P)\n",
      "\n",
      "print (\"(ii) Specific volume and density\")\n",
      "SV = V/m; \n",
      "print \"Specific volume = %.3f m**3/kg\"%(SV)\n",
      "\n",
      "D = m/V;\n",
      "print \"Density = %.3f kg/m**3\"%(D),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) Total pressure in the vessel\n",
        "P = 1.841 bar\n",
        "(ii) Specific volume and density\n",
        "Specific volume = 0.037 m**3/kg\n",
        "Density = 27.162 kg/m**3\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.6 page no : 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Calculate the gas pressure.\n",
      "'''\n",
      "\n",
      "\n",
      "# Variables\n",
      "h0 = .761; \t\t\t#m\n",
      "h = .55; \t\t\t#m\n",
      "g = 9.79; \t\t\t#m/s**2\n",
      "rho = 13640.; \t\t\t#kg/m**3\n",
      "\n",
      "# Calculations\n",
      "P = rho*g*(h0+h); \t\t\t#N/m**2\n",
      "\n",
      "# Results\n",
      "print \"Gas pressure = %.3f bar\"%(P/10**5),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Gas pressure = 1.751 bar\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.7 page no : 44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Find the absolute pressure of steam.\n",
      "'''\n",
      "\n",
      "# Variables\n",
      "h_H2O = 34.; \t\t\t#mm of Hg\n",
      "g = 9.81; \t\t\t#m/s**2\n",
      "rho = 13600.; \t\t\t#kg/m**3\n",
      "P_Hg = 97.5; \t\t\t#mm of Hg\n",
      "P_atm = 760.; \t\t\t#mm of Hg\n",
      "\n",
      "# Calculations\n",
      "P_H2O = h_H2O/13.6; \t\t\t#mm of Hg\n",
      "Pabs = rho*g*(P_Hg+P_atm-P_H2O)/10**8; \t\t\t#bar\n",
      "\n",
      "# Results\n",
      "print \"absolute pressure  = %.3f bar\"%(Pabs)\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "absolute pressure  = 1.141 bar\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.8 page no : 44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Find the absolute pressure of the gas \n",
      "'''\n",
      "\n",
      "# Variables\n",
      "SG = 0.8;\n",
      "rho_H2O = 1000.; \t\t\t#kg/m**3\n",
      "g = 9.81; \t\t\t#ms**2\n",
      "h = 0.17; \t\t\t#m\n",
      "Patm = 1.01325; \t\t\t#bar\n",
      "\n",
      "# Calculations\n",
      "rho = SG*rho_H2O; \t\t\t#kg/m**3\n",
      "P_liq = rho*g*h/10**5; \t\t\t#bar\n",
      "P_gas = Patm - P_liq;\n",
      "\n",
      "# Results\n",
      "print \"gas pressure =  %.3f bar\"%(P_gas),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "gas pressure =  1.000 bar\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.9 page no : 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Estimate the mass of a piston that can be supported by a gas entrapped\n",
      "under the piston '''\n",
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "d = 0.2; \t\t\t#m\n",
      "g = 9.81; \t\t\t#m/s**2\n",
      "h = 0.117; \t\t\t#m\n",
      "rho = 13600.; \t\t\t#kg/m**3\n",
      "\n",
      "# Calculations\n",
      "p = rho*g*h;\n",
      "m = (p*math.pi/4*d**2)/g;\n",
      "\n",
      "# Results\n",
      "print \"mass = %.3f kg\"%(m),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "mass = 49.989 kg\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.10 page no : 49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "calculate the kinetic energy of an artificial satellite revolves round the earth \n",
      "'''\n",
      "\n",
      "# Variables\n",
      "v = 800.; \t\t\t#m/s\n",
      "g = 9.; \t\t\t#m/s**2\n",
      "F = 3600.; \t\t\t#N\n",
      "\n",
      "# Calculations\n",
      "m = F/g;\n",
      "KE = 1./2*m*v**2./10**6;\n",
      "\n",
      "# Results\n",
      "print \"Kinetic Energy = %.3f MJ\"%(KE),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Kinetic Energy = 128.000 MJ\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.11 page no : 49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "find :\n",
      "(i) Heat transferred ;\n",
      "(ii) Mean specific heat of the gas.\n",
      "'''\n",
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "# Variables\n",
      "m = 6.; \t\t\t#kg\n",
      "T1 = 25.; \t\t\t#0C\n",
      "T2 = 125.; \t\t\t#0C\n",
      "\n",
      "print (\"(i) Heat transferred\")\n",
      "\n",
      "# Calculations and Results\n",
      "def f18(T): \n",
      "\t return m*(0.4+0.004*T)\n",
      "\n",
      "Q =  quad(f18,T1,T2)[0]\n",
      "\n",
      "print \"heat tranferred = %.3f kJ\"%(Q)\n",
      "\n",
      "print (\"(ii) Mean specific heat of the gas\")\n",
      "c_n = Q/m/(T2-T1);\n",
      "print \"Mean specific heat = %.3f kJ/kg.0C\"%(c_n),\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) Heat transferred\n",
        "heat tranferred = 420.000 kJ\n",
        "(ii) Mean specific heat of the gas\n",
        "Mean specific heat = 0.700 kJ/kg.0C\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.12 page no : 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "what will be the temperature corresponding to the thermometric property of 3.5 on\n",
      "Celsius scale.\n",
      "'''\n",
      "\n",
      "import math \n",
      "from numpy import *\n",
      "# Variables\n",
      "Ice_point = 0.;\n",
      "Steam_point = 100.;\n",
      "\n",
      "# Calculations\n",
      "P = [[math.log(1.5),1],[math.log(7.5),1]];\n",
      "Q = [0,100];\n",
      "X = linalg.inv(P)*Q;\n",
      "\n",
      "a = X[0,1];\n",
      "b = X[1,1];\n",
      "p = 3.5;\n",
      "t = a*math.log(p)+b;\n",
      "\n",
      "# Results\n",
      "print (\"The value of temperature is given by %.3f\")%(t),(\"\u00b0C\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of temperature is given by 52.646 \u00b0C\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.13 page no : 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "What will be the reading on the thermometer where the gas thermometer reads 70\u00b0C ?\n",
      "'''\n",
      "\n",
      "def func(t): \n",
      "\t return 0.20*t-5*10**(-4)*t**2\n",
      "# Variables # Calculations\n",
      "t1 = 0; \t\t\t#0C\n",
      "e1 = func(t1);\n",
      "t2 = 100; \t\t\t#0C\n",
      "e2 = func(t2);\n",
      "t3 = 70; \t\t\t#0C\n",
      "e3 = func(t3);\n",
      "t = e3*(t2-t1)/e2-e1;\n",
      "\n",
      "# Results\n",
      "print \"thermocouple will read\",t,\"\u00b0C\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "thermocouple will read 77.0 \u00b0C\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.15 page no : 51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "What is the amount of work done upon the atmosphere by the balloon ? \n",
      "'''\n",
      "\n",
      "# Variables\n",
      "p = 101.325; \t\t#kPa\n",
      "V2 = 0.6; \t\t\t#m**3\n",
      "V1 = 0; \t\t\t#m**3\n",
      "\n",
      "# Calculations\n",
      "W = p*(V2-V1);\n",
      "\n",
      "# Results\n",
      "print (\"work done by atmosphere = \"),(-W),\"kJ\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "work done by atmosphere =  -60.795 kJ\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.16 page no : 52"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Determine the work done by the air which enters into an evacuated vessel\n",
      "from atmosphere when the valve is opened'''\n",
      "\n",
      "from scipy import integrate\n",
      "\n",
      "# Variables\n",
      "#p = 1.013*10**5; \t#N/m**2\n",
      "p = lambda x: 1.013*10**5\n",
      "V1 = 1.5; \t\t\t#m**3\n",
      "V2 = 0; \t\t\t#m**3\n",
      "\n",
      "# Calculations\n",
      "integ, err = integrate.quad(p,V1,V2)\n",
      "\n",
      "# Results\n",
      "print (\"W = \"),(integ/10**3),\"kJ\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "W =  -151.95 kJ\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.17 page no : 53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "A piston and cylinder machine containing a fluid system has a stirring\n",
      "device as shown in Fig. 2.36. Find the net work transfer for the system.\n",
      "'''\n",
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "T = 1.25; \t\t\t#N.m\n",
      "N = 9500.;\n",
      "p = 101.3; \t\t\t#kPa\n",
      "d = 0.65; \t\t\t#m\n",
      "L = 0.6; \t\t\t#m\n",
      "\n",
      "# Calculations\n",
      "W1 = 2*math.pi*N*T/1000; \t#kJ\n",
      "A = math.pi/4*d**2; #m**2\n",
      "W2 = p*A*L; \t\t#kJ\n",
      "Wnet = (-W1)+W2;\n",
      "\n",
      "# Results\n",
      "print \"The net work transfer for the system = %.3f\"%(Wnet),\"kJ\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The net work transfer for the system = -54.444 kJ\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.18 page no : 53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Find the net work done during the process.\n",
      "'''\n",
      "\n",
      "# Variables\n",
      "A = 45.*10**(-4); \t  #m**2\n",
      "P = 0.9*10**5; \t\t  #N/m**2\n",
      "Patm = 1.013*10**5;   #N/m**2\n",
      "L = 0.05; \t\t\t  #m\n",
      "\n",
      "# Calculations\n",
      "dV = 300.*10**(-6); \t\t\t#m**3\n",
      "W = P*A*L-Patm*dV;\n",
      "\n",
      "# Results\n",
      "print (\"net work done  = \"),(W),\"J\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "net work done  =  -10.14 J\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.19 page no : 54"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Calculate the work done when the pressure increases from 1.5 bar to 7.5 bar.\n",
      "'''\n",
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "# Variables\n",
      "p1 = 1.5; \t\t\t#bar\n",
      "p2 = 7.5; \t\t\t#bar\n",
      "V1 = 3/p1;\n",
      "V2 = 3/p2;\n",
      "\n",
      "# Calculations\n",
      "def f19( V): \n",
      "\t return 3./V*10**2\n",
      "\n",
      "W =  quad(f19, V1, V2)[0]\n",
      "\n",
      "# Results\n",
      "print \"Work done = %.3f\"%(W),\"kJ\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Work done = -482.831 kJ\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.20 page no : 55"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "'''\n",
      "To a closed system 150 kJ of work is supplied, determine the final volume and pressure of the system.\n",
      "'''\n",
      "\n",
      "# Variables\n",
      "W = 150; \t\t\t#kJ\n",
      "V1 = 0.6; \t\t\t#m**3\n",
      "\n",
      "# Calculations and Results\n",
      "V2 = (8-math.sqrt(64-4*2*2.58))/4; \t\t\t#m**3\n",
      "print (\"Final volume  = %.3f\")%V2,\"m**3\"\n",
      "\n",
      "p2 = 8-4*V2;\n",
      "print \"Final pressure  = %.2f\"%p2,\"bar\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Final volume  = 0.354 m**3\n",
        "Final pressure  = 6.58 bar\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.21 page no : 56"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Calculate the work done by the fluid on the piston.\n",
      "'''\n",
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "# Variables\n",
      "p1 = 3.*10**5; \t\t\t#Pa\n",
      "v1 = 0.18; \t\t\t#m**3/kg\n",
      "p2 = 0.6*10**5; \t\t\t#Pa\n",
      "\n",
      "# Calculations\n",
      "C = p1*v1**2;\n",
      "v2 = math.sqrt(C/p2);\n",
      "\n",
      "def f17( v): \n",
      "    return C/v**2\n",
      "\n",
      "W =  quad(f17, v1,v2)[0]\n",
      "\n",
      "# Results\n",
      "print (\"Work done = %d\")%(W),(\"Nm/kg\")\n",
      "\n",
      "# Note : output would be differ as rounding error is there. v2 has rounding off error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Work done = 29850 Nm/kg\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.22 page no : 57"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "'''\n",
      "Calculate the net work done by the fluid, for an initial volume of 0.05 m 3 .\n",
      "'''\n",
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "# Variables\n",
      "m = 1.; \t\t\t#kg\n",
      "p1 = 20.*10**5; \t#Pa\n",
      "V1 = 0.05; \t\t\t#m**3\n",
      "\n",
      "# Calculations\n",
      "V2 = 2*V1;\n",
      "p2 = p1*(V1/V2)**2;\n",
      "C = p1*V1**2;\n",
      "V3 = V1;\n",
      "\n",
      "def f20( V): \n",
      "\t return C/V**2\n",
      "\n",
      "W_12 =  quad(f20, V1,V2)[0]\n",
      "W_23 = p2*(V2-V3);\n",
      "W_net = W_12-W_23;\n",
      "\n",
      "# Results\n",
      "print (\"Net work done  =  \"),(W_net),(\"Nm\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Net work done  =   25000.0 Nm\n"
       ]
      }
     ],
     "prompt_number": 26
    }
   ],
   "metadata": {}
  }
 ]
}