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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Ch-6, Unconventional Machining Process"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 1 on page no. 332"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi, sqrt\n",
      "from __future__ import division\n",
      "# Given that\n",
      "a = 5 # Side of the square hole in mm\n",
      "t = 4 # Thickness of tungsten plate in mm\n",
      "d = 0.01 # Diameter of abraisive grains in mm\n",
      "F = 3.5 # Force for feeding in N\n",
      "A =25e-3 # Amplitude of tool oscillation in mm\n",
      "f = 25e3 # Frequency in Hz\n",
      "Hw = 6900 # Fracture hardness of WC in N/mm**2\n",
      "\n",
      "Z = (1/2)*(4*a**2)/(pi*d**2)\n",
      "lamda = 5\n",
      "d1 = (d**2)\n",
      "h_w = (sqrt((8*F*A)/(pi*Z*d1*Hw*(1+lamda))))\n",
      "Q = (2/3)*((d1*h_w)**(3/2))*Z*f*pi\n",
      "t = (a**2)*t/(Q*60)\n",
      "print \"The approximate time required = %0.2f min\"%(t)\n",
      "# Answer in the book is given as 13.66 min"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The approximate time required = 14.26 min\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 2 on page no. 334"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "r = 1/3 # Ratio of hardness values of copper and steel\n",
      "R_Q = (r)**(3/4)\n",
      "R_t = 1/R_Q\n",
      "P_R = (1-(1/R_t))*100\n",
      "print \"Percentage change in cutting time when tool is changed from coppper to steel = %d percent(reduction)\"%(P_R)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Percentage change in cutting time when tool is changed from coppper to steel = 56 percent(reduction)\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 3 on page no. 345"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "m = 5 # Romoval rate in cm**3/min\n",
      "A = 56 # Atomic gram weight in gm\n",
      "Z = 2 # Valence at which dissolation takes place\n",
      "D = 7.8 # Density of iron in gm/cm**3\n",
      "I = (m/60)*(D*Z*96500)/(A)\n",
      "print \"Current required = %d amp\"%(I)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Current required = 2240 amp\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 4 on page no. 345"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "I = 1000 # Current in amp\n",
      "p1 = 72.5 # Percentage(by weight) of Ni in Nimonic 75 alloy\n",
      "p2 = 19.5 # Percentage(by weight) of Cr in Nimonic 75 alloy\n",
      "p3 = 5 # Percentage(by weight) of Fe in Nimonic 75 alloy\n",
      "p4 = 0.4 # Percentage(by weight) of Ti in Nimonic 75 alloy\n",
      "p5 = 1 # Percentage(by weight) of Si in Nimonic 75 alloy\n",
      "p6 = 1 # Percentage(by weight) of Mn in Nimonic 75 alloy\n",
      "p7 = 6 # Percentage(by weight) of Cu in Nimonic 75 alloy\n",
      "# From the table 6.3 given in the book\n",
      "D1 = 8.9 # Density of Ni in g/cm**3\n",
      "D2 = 7.19 # Density of Cr in g/cm**3\n",
      "D3 = 7.86 # Density of Fe in g/cm**3\n",
      "D4 = 4.51 # Density of Ti in g/cm**3\n",
      "D5 = 2.33 # Density of Si in g/cm**3\n",
      "D6 = 7.43 # Density of Mn in g/cm**3\n",
      "D7 = 8.96 # Density of Cu in g/cm**3\n",
      "A1 = 58.71 # Gram atomic weight of Ni in gm\n",
      "A2 = 51.99 # Gram atomic weight of Cr in gm\n",
      "A3 = 55.85 # Gram atomic weight of Fe in gm\n",
      "A4 = 47.9 # Gram atomic weight of Ti in gm\n",
      "A5 = 28.09 # Gram atomic weight of Si in gm\n",
      "A6 = 54.94 # Gram atomic weight of Mn in gm\n",
      "A7 = 63.57 # Gram atomic weight of Cu in gm\n",
      "Z1 = 2 # Valence of dessolation for Ni\n",
      "Z2 = 2 # Valence of dessolation for Cr\n",
      "Z3 = 2 # Valence of dessolation for Fe\n",
      "Z4 = 3 # Valence of dessolation for Ti\n",
      "Z5 = 4 # Valence of dessolation for Si\n",
      "Z6 = 2 # Valence of dessolation for Mn\n",
      "Z7 = 1 # Valence of dessolation for Cu\n",
      "# Above values are given in table 6.3 in the book\n",
      "D = 100/((p1/D1)+(p2/D2)+(p3/D3)+(p4/D4)+(p5/D5)+(p6/D6)+(p7/D7))\n",
      "Q = ((0.1035*(10**-2))/D)*(1/((p1*Z1/A1)+(p2*Z2/A2)+(p3*Z3/A3)+(p4*Z4/A4)+(p5*Z5/A5)+(p6*Z6/A6)+(p7*Z7/A7)))\n",
      "R = Q*I*60\n",
      "print \"Removal rate = %0.1f cm**3/min\"%(R)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Removal rate = 2.2 cm**3/min\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 5 on page no. 352"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "V = 10 # DC supply voltage in Volt\n",
      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
      "f = 0.1 # Feed rate in m/min\n",
      "Vo = 1.5 # Total overvoltage in Volt\n",
      "F = 96500 # Faraday constant in coulombs per mole\n",
      "\n",
      "A = 55.85 # Atomic gram weight of iron in gm\n",
      "Z = 2 # Valency of dissolation of iron\n",
      "rho = 7.86 # Density of iron in gm/cm**3\n",
      "Yc = k*A*(V-Vo)/(rho*Z*F*(f/60))\n",
      "print \"Equilibrium gap = %0.2f cm\"%(Yc)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Equilibrium gap = 0.04 cm\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 6 on page no. 353"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "S_I1 = 5 # Surface irregulation in micro meter\n",
      "S_I2 = 8 # Surface irregulation in micro meter\n",
      "V = 12 # DC supply voltage in Volt\n",
      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
      "Vo = 1.5 # Total overvoltage in Volt\n",
      "F = 96500 # Faraday constant in coulombs per mole\n",
      "Y_min = (S_I1+S_I2)*(10**(-4))\n",
      "A = 55.85 # Atomic gram weight of iron in gm\n",
      "Z = 2 # Valency of dissolation of iron\n",
      "D = 7.86 # Density of iron in gm/cm**3\n",
      "f_max = (k*A*(V-Vo)/(Z*D*F*Y_min))*60\n",
      "print \"Largest passible feed rate = %0.1f mm/min\"%(f_max*10)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Largest passible feed rate = 35.7 mm/min\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 7 on page no. 355"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "f  = 0.2 # Feed rate in cm/min\n",
      "l = 2.54 # Length of tool face in cm\n",
      "w = 2.54 # Width of tool face in cm\n",
      "T_b = 95 # Boiling temperature of electrolyte in \u00b0C\n",
      "Nita = 0.876e-3 # Viscosity of electrolyte in kg/m-sec\n",
      "D_e = 1.088 # Density of electrolyte in g/cm**3\n",
      "c = .997 # Specific heat of electrolyte\n",
      "V = 10 # DC supply voltage in Volt\n",
      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
      "T = 35 # Ambient temperature in \u00b0C\n",
      "Vo = 1.5 # Total overvoltage in Volt\n",
      "F = 96500 # Faraday constant in coulombs per mole\n",
      "A = 55.85 # Atomic gram weight of iron in gm\n",
      "Z = 2 # Valency of dissolation of iron\n",
      "D = 7.86 # Density of iron in gm/cm**3\n",
      "Ye = k*A*(V-Vo)*60/(D*Z*F*f)\n",
      "J = k*(V-Vo)/(Ye)\n",
      "D_T = T_b -T\n",
      "v = (J**2)*(l)/(k*D_T*D_e*c)\n",
      "Re = ((D_e*v*2*Ye)/Nita)*(0.1)\n",
      "p = 0.3164*D_e*(v**2)*l/(4*Ye*(Re**0.25))*(10**-4)\n",
      "A = l*w\n",
      "F = p*A*(10**-1)*(1/2)\n",
      "print \"Total force acting on the tool = %d N\"%(F)\n",
      "# Answer in the book is given as 79 N"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total force acting on the tool = 103 N\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 9 on page no. 378"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "# Given that\n",
      "a = 10 # Side length of a square hole in mm\n",
      "t = 5 # Thickness of low carbon steel plate in mm\n",
      "R = 50 # Resistance in relaxation circuit in ohm\n",
      "C = 10 # Capacitance in relaxation circuit in micro F\n",
      "V = 200 # Supply voltage in Volt\n",
      "V_ = 150 # Minimum required voltage for discharge in Volt\n",
      "\n",
      "E = (1/2)*C*(10**-6)*(V_**2)\n",
      "tc = R*C*(10**-6)*log(V/(V-V_))\n",
      "W = (E/tc)*(10**-3)\n",
      "v = t*a**2\n",
      "Q = 27.4*(W**(1.54))\n",
      "T = v/Q\n",
      "print \"The time required to complete the drilling operation = %d min\"%(T)\n",
      "# Answer in the book is given as 306 min"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The time required to complete the drilling operation = 300 min\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 10 on page no. 382"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "R = 50 # Resistance in relaxation circuit in ohm\n",
      "C = 10 # Capacitance in relaxation circuit in micro F\n",
      "V = 200 # Supply voltage in Volt\n",
      "V_ = 150 # Minimum required voltage for discharge in Volt\n",
      "E = (1/2)*C*(10**-6)*(V_**2)\n",
      "tc = R*C*(10**-6)*log(V/(V-V_))\n",
      "W = (E/tc)*(10**-3)\n",
      "Q = 27.4*(W**(1.54))\n",
      "Hrms = 1.11*(Q**0.384)\n",
      "print \"Surface roughness = %0.3f micro meter\"%(Hrms)\n",
      "# Answer in the book is given as 5.16 micro meter which is wrong"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Surface roughness = 1.350 micro meter\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 11 on page no. 391"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "w = 150 # Width of slot in micro meter\n",
      "t = 1 # Thickness of tungsten sheet in mm\n",
      "P = 5 # Power of electron beam in KW\n",
      "C = 12 # Specific power consumption for tugsten in W/(mm**3/min) from the table 6.7 given in the book \n",
      "v = (P*(1000)/C)*(1000/(w*t))*(1/600)\n",
      "print \"Speed of cutting = %0.1f cm/sec\"%(v)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Speed of cutting = 4.6 cm/sec\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 12 on page no. 392"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import ceil\n",
      "# Given that\n",
      "V = 150e3 # Acceleration voltage in V\n",
      "D = 76e-7 # Density of steel in kg/mm**3\n",
      "Delta = 2.6*(10**-17)*((V**2)/D)\n",
      "print \"Electron range = %d micro meter\"%(ceil(Delta*(10**3)))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Electron range = 77 micro meter\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 13 on page no. 395"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "w = 0.015 # Width of slot in cm\n",
      "t = 1 # Thickness of tungsten sheet in mm\n",
      "P = 5e3 # Power of electron beam in W\n",
      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
      "T_m = 3400 # Melting temperture in \u00b0C\n",
      "Z = t/10 # In cm\n",
      "v = (0.1**2)*(P**2)/((T_m**2)*(Z**2)*(k*w*rho_c))\n",
      "print \"Speed of cutting = %0.1f cm/sec\"%(v)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Speed of cutting = 24.7 cm/sec\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 14 on page no. 399"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
      "p_a = 10 # Percentage of beam absorbed\n",
      "alpha = k/rho_c\n",
      "H = (p_a/100)*(I)*(100)\n",
      "tm = (pi/alpha)*((T_m*k)/(2*H))**(2)\n",
      "print \"Time required for the surface to reach the melting point = %f sec\"%(tm)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time required for the surface to reach the melting point = 0.000053 sec\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 15 on page no. 400"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
      "d = 200 # Focused diameter of incident beam in micro meter\n",
      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
      "p_a = 10 # Percentage of beam absorbed\n",
      "H = (p_a/100)*(I)*(100)\n",
      "alpha = k/rho_c\n",
      "zeta = 0.5 # Fr0m the standard table\n",
      " # By solving the equation T_m = ((2*H)*(sqrt(alpha*tm))/k)*((1/sqrt(pi))-ierfc(d/(4*sqrt(alpha*tm))))\n",
      "tm = 1/((200**2)*(zeta**2)*(alpha))\n",
      "print \"Time required for the centre of the circular spot to reach the melting point = %f sec\"%(tm)\n",
      "# Answer in the book is given as 0.00013 sec"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time required for the centre of the circular spot to reach the melting point = 0.000126 sec\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 16 on page no. 401"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "d = 200 # Diameter of focussed laser beam in micro meter\n",
      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
      "p_a = 10 # Percentage of beam absorbed\n",
      "H = 2*k*T_m/(d*10**-4)\n",
      "I = H/(p_a/100)\n",
      "print \"Minimum value of beam power intensity to achieve the melting = %0.2e W/cm**2\"%(I)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Minimum value of beam power intensity to achieve the melting = 7.31e+06 W/cm**2\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 17 on page no. 403"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
      "t = 0.5 # Thickness of tungsten sheet in mm\n",
      "d = 200 # Drill diameter in micro meter\n",
      "P = 3e4 # Energy required per unit volume to vapourize tungsten in J/cm**3\n",
      "p_e = 10 # Percentage efficiency\n",
      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
      "H = (p_e/100)*(I)*(100)\n",
      "v = H/P\n",
      "T = t*(0.1)/(v)\n",
      "print \"The time required to drill a through hole = %0.4f sec\"%(T)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The time required to drill a through hole = 0.0015 sec\n"
       ]
      }
     ],
     "prompt_number": 16
    }
   ],
   "metadata": {}
  }
 ]
}