path: root/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2_2.ipynb

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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Ch-2, Casting Processes"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 1 on page no. 46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt\n",
      "# Given that\n",
      "h=15 # Height of spur in cm\n",
      "l= 50 # Length of cast in cm\n",
      "w= 25 # weidth of cast in cm\n",
      "h1= 15 # Height of cast in cm\n",
      "g= 981 # Acceleration due to gravity in cm/sec**2\n",
      "Ag= 5 # Cross sectional area of the grate in cm**2\n",
      "v3= sqrt(2* g * h)\n",
      "V = l*w*h1\n",
      "tf1= V/(Ag*v3)\n",
      "Am = l*w\n",
      "tf2 = (Am/Ag)*(1/sqrt(2*g))*2*(sqrt(h) - sqrt(h-h1))\n",
      "print \"Filling time for first design = %0.2f sec, \\nFilling time for second design = %0.2f sec\"%(tf1, tf2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Filling time for first design = 21.86 sec, \n",
        "Filling time for second design = 43.72 sec\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 2 on page no. 53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt, pi\n",
      "# Given that\n",
      "h=15 # Height of spur in cm\n",
      "l= 50 # Length of cast in cm\n",
      "w= 25 # weidth of cast in cm\n",
      "h1= 15 # Height of cast in cm\n",
      "g= 981 # Acceleration due to gravity in cm/sec**2\n",
      "Ag= 5 # Cross sectional area of the grate in cm**2\n",
      "Dm = 7800 # Density of molten Fe in Kg/m**3\n",
      "Neta = 0.00496 # Kinetic viscosity in Kg/m-sec\n",
      "theta = 90 # Angle in degree\n",
      "Eq = 25 # (L/D) Equivalent \n",
      "\n",
      "v3= sqrt(2* g * h)*(10**(-2))\n",
      "d= sqrt((Ag*4)/(pi))*(10**(-2))\n",
      "Re = Dm*v3*d/Neta\n",
      "f = 0.0791*(Re)**(-1/4)\n",
      "L=0.12 # in meter\n",
      "Cd= (1+0.45+4*f*((L/d)+Eq))**(-1/2)\n",
      "v3_ = Cd*v3\n",
      "Re_ = (v3_/v3)*(Re)\n",
      "f_ = 0.0791 *(Re_)**(-1/4)\n",
      "Cd_ = (1+0.46+4*f_*(L/d + Eq))**(-1/2)\n",
      "v3__ = Cd_*v3\n",
      "V = l*w*h1\n",
      "tf= (V/(Ag*v3__))*(10**-2)\n",
      "print \"Filling time for first design = %0.1f sec. \"% tf"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Filling time for first design = 31.7 sec. \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 3 on page no. 55"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt, pi\n",
      "# Given that\n",
      "Hi=1.2 # Initial height in m\n",
      "H= 0.05 # Height in m\n",
      "g= 9.81 # Acceleration due to gravity in m/sec**2\n",
      "Dm = 2700 # Density of molten metal in Kg/m**3\n",
      "Neta = 0.00273 # Kinetic viscosity in Kg/m-sec \n",
      "d= 0.075 # Diameter in m\n",
      "D = 1 # Internal diameter of ladle in m\n",
      "\n",
      "v3= sqrt(2* g * Hi)\n",
      "Re = Dm*v3*d/Neta\n",
      "ef=0.075\n",
      "Cd= (1+ef)**(-1/2)\n",
      "ef_=0.82\n",
      "Re_ = (2+ef_)**(-1/2)\n",
      "v3_ = sqrt(2*g*H)\n",
      "Re_ = Dm*v3_*d/Neta\n",
      "At = (pi/4)*D**2\n",
      "An = (pi/4)*d**2\n",
      "Cd= 0.96\n",
      "tf= (sqrt(2/g))*(At/An)*(1/Cd)*sqrt(Hi)\n",
      "m = Dm*An*Cd*sqrt(2*g*Hi)\n",
      "m_ = Dm*An*Cd*sqrt(2*g*Hi*0.25)\n",
      "print \"\"\"Time required to empty the ladle = %0.2f sec,\n",
      "Discharge rate are - \n",
      "  Initially = %0.2f Kg/sec \n",
      "  When the ladle is 75 percent empty = %0.2f Kg/sec.\"\"\"%(tf,m,m_)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time required to empty the ladle = 91.60 sec,\n",
        "Discharge rate are - \n",
        "  Initially = 55.56 Kg/sec \n",
        "  When the ladle is 75 percent empty = 27.78 Kg/sec.\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 5 on page no. 66"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt, pi\n",
      "# Given that\n",
      "thetaF= 1540 # Temperature of mould face in degree centigrade\n",
      "Theta0 = 28 # Initial temperature of mould in Degree centigrade\n",
      "L= 272e3 # Latent heat of liquid metal in J/Kg\n",
      "Dm = 7850 # Density of liquid metal in Kg/m**3\n",
      "c = 1.17e+3 #Specific heat of sand in J/Kg-K\n",
      "k = 0.8655 # Conductivity of sand in W/m-K\n",
      "D= 1600 # Density of sand in Kg/m**3\n",
      "h = 0.1 # Height in m\n",
      "b = 10 # Thickness of slab in cm\n",
      "r =h/2# V/A in meter\n",
      "\n",
      "lamda = (thetaF - Theta0)*(D*c)/(Dm*L)\n",
      "Beta1 = 2*lamda/sqrt(pi)\n",
      "Alpha = k /(D*c)\n",
      "ts1 = r**2 /((Beta1**2)*Alpha)#In sec\n",
      "ts1_=ts1/3600 # In hour\n",
      "from sympy import symbols, solve\n",
      "Beta= symbols(\"Beta\") \n",
      "p=Beta**2 - lamda*(2/sqrt(pi))*Beta -lamda/3\n",
      "Beta2 = solve(p, Beta) # taking +ve value\n",
      "print \"Beta2 = %0.2f\" %Beta2[1]\n",
      "r1 = r/3\n",
      "ts2 = (r1**2)/((1.75**2)*Alpha) # in sec\n",
      "ts2_=ts2/3600#in Hour\n",
      "print \"Solidification time for slab-shaped casting = %0.3f hr,\\nSolidification time for sphere = %0.3f hr\"% (ts1_,ts2_)\n",
      "# Answer in the textbook are not accurate."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Beta2 = 1.75\n",
        "Solidification time for slab-shaped casting = 0.671 hr,\n",
        "Solidification time for sphere = 0.054 hr\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 6 on page no. 73"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt, pi, exp\n",
      "# Given that\n",
      "thetaF= 1540 # Temperature of mould face in degree centigrate\n",
      "ThetaO = 28 # Initial temperature of mould in Degree centigrate\n",
      "L= 272e3 # Latent heat of iron in J/Kg\n",
      "Dm = 7850 # Density of iron in Kg/m**3\n",
      "Cs = 0.67e+3 #Specific heat of iron in J/Kg-K\n",
      "C = 0.376e3 #Specific heat of copper in J/Kg-K\n",
      "Ks = 83 # Conductivity of iron in W/m-K\n",
      "K = 398 # Conductivity of copper in W/m-K\n",
      "D= 8960 # Density of copper in Kg/m**3\n",
      "h = .1 # Height in m\n",
      "\n",
      "zeta1=0.98#By solving eqauation- zeta*exp(zeta**2)*erf(zeta)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.98\n",
      "AlphaS = Ks /(Dm*Cs)\n",
      "ts1 = h**2 / (16*(zeta1**2) * AlphaS)#In sec\n",
      "ts1_=ts1/3600 # In hour\n",
      "Phi = sqrt((Ks*Dm*Cs)/(K*D*C))\n",
      "zeta2=0.815#By solving eqauation- zeta*exp(zeta**2)*(erf(zeta)+Phi)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.815\n",
      "ts2 = h**2 / (16*(zeta2**2) * AlphaS)#In sec\n",
      "ts2_=ts2/3600 # In hour\n",
      "erf = lambda zeta2:(thetaF-Theta0)*Cs/L/sqrt(pi)\n",
      "thetaS= thetaF+(thetaF-(L*(sqrt(pi))*zeta2*(exp(zeta2**2))*erf(zeta2))/Cs)\n",
      "print \"\"\"Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = %0.4f hr,\n",
      "Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = %0.4f hr,\n",
      "The surface temperature of the mould = %0.1f\u00b0 C\"\"\"% (ts1_,ts2_,thetaS)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = 0.0115 hr,\n",
        "Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = 0.0166 hr,\n",
        "The surface temperature of the mould = 685.7\u00b0 C\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 7 on page no. 75"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import sqrt, pi, exp\n",
      "# Given that\n",
      "thetaF= 1540 # Temperature of mould face in degree centigrade\n",
      "thetaO = 28 # Initial temperature of mould in Degree centigrade\n",
      "L= 272e3 # Latent heat of iron in J/Kg\n",
      "Dm = 7850 # Density of iron in Kg/m**3\n",
      "Cs = 0.67e+3 #Specific heat of iron in J/Kg-K\n",
      "C = 0.376e3 #Specific heat of copper in J/Kg-K\n",
      "Ks = 83 # Conductivity of iron in W/m-K\n",
      "K = 398 # Conductivity of copper in W/m-K\n",
      "D= 8960 # Density of copper in Kg/m**3\n",
      "h = .1 # Height in m\n",
      "hF = 1420 # Total heat transfer coefficient across the casting-mould interface in W/m**2-\u00b0C\n",
      "\n",
      "AlphaS = K /(D*C)\n",
      "thetaS = 982 #In \u00b0C as in example 2.6\n",
      "h1= (1+(sqrt((Ks*Dm*Cs)/(K*D*C))))*hF\n",
      "a = 1/2 + (sqrt((1/4)+Cs*(thetaF-thetaS)/(3*L)))\n",
      "delta=h/2\n",
      "ts = (delta+((h1*delta**2)/(2*Ks)))/((h1*(thetaF-thetaS))/(Dm*L*a)) # in sec\n",
      "ts_ = ts/3600 # in hours\n",
      "h2= (1+(sqrt((K*D*C)/(Ks*Dm*Cs))))*hF\n",
      "gama= ((h2**2)/(K**2))*AlphaS*ts\n",
      "thetaS_ = thetaO + (thetaS-thetaO)*(1-((exp(gama))*(1-(erf(sqrt(gama))))))\n",
      "print \"Solidification time = %0.4f hr,\\nThe surface temperature of the mould = %0.2f \u00b0 C\"%(ts_,thetaS_)\n",
      "# The value of the surface temperature of the mould in the book is given as 658.1\u00b0 C, Which is wrong."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Solidification time = 0.0534 hr,\n",
        "The surface temperature of the mould = 10389.84 \u00b0 C\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 8 on page no. 77"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given that\n",
      "A= 60*7.5 # Cross sectional area in cm**2\n",
      "v=0.05 # Withdrawal rate in m/sec\n",
      "t = 0.0125 # Thickness in m\n",
      "thetaF= 1500 # Temperature of mould face in degree centigrate\n",
      "thetaP = 1550 # \n",
      "thetaO = 20 # Initial temperature of mould in Degree centigrate\n",
      "L= 268e3 # Latent heat of molten metal in J/Kg\n",
      "Dm = 7680 # Density of molten metal in Kg/m**3\n",
      "Cs = 0.67e+3 #Specific heat of molten metal in J/Kg-K\n",
      "Cm = 0.755e3 #Specific heat of mould in J/Kg-K\n",
      "Ks = 76 # Conductivity of molten metal in W/m-K\n",
      "hF = 1420 # Heat transfer coefficient at the casting-mould interface in W/m**2-\u00b0C\n",
      "Dtheta = 10 # Maximum temperature of cooling water in \u00b0 C\n",
      "L_ = L+Cm*(thetaP-thetaF)\n",
      "x=L_ / (Cs*(thetaF-thetaO))\n",
      "y= hF*t/Ks\n",
      "print \"L_/(Cs(thetaF-thetaO))=%0.2f,\\nhF*t/Ks=%0.2f\"%(x,y)\n",
      "z=0.11 # Where z=hF**2 * lm / (v*Ks*Dm*Cs)\n",
      "lm= (z*v*Ks*Dm*Cs)/(hF**2)\n",
      "Z=0.28 # Where Z=Q/(lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks))\n",
      "Q = Z*lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks)\n",
      "m = Q / (4.2e3*Dtheta)\n",
      "print \"The mould length = %0.2f meter,\\nThe cooling water requirement = %0.2f Kg/sec\"%(lm,m)\n",
      "# Answer for The cooling water requirement in the book is given as 5.05 Kg/sec, Which is wrong."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "L_/(Cs(thetaF-thetaO))=0.31,\n",
        "hF*t/Ks=0.23\n",
        "The mould length = 1.07 meter,\n",
        "The cooling water requirement = 48.07 Kg/sec\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem 9 on page no. 81"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import floor\n",
      "# Given that\n",
      "a = 15 # Side of the aluminium cube in cm\n",
      "Sh = 0.065 # Volume shrinkage of aluminium during solidification\n",
      "\n",
      "Vc = a**3\n",
      "Vr = 3*Sh*Vc\n",
      "h = ((4*Vr)/pi)**(1/3)\n",
      "Rr = 6/h # Where Rr= (A/V)r\n",
      "Rc = 6/a # Where Rc = (A/V)c\n",
      "print \"(A/V)r=%f, (A/V)c=%.1f\\nHence Rr is greater than Rc\"%(Rr,Rc)\n",
      "dmin = 6/Rc\n",
      "Vr_ = (pi/4)*dmin**3\n",
      "print \"\"\"With minimum value of d Vr=%d cm**3 .\n",
      "This valume is much more than the minimum Vr necessary.\n",
      "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2\n",
      "and again (A/V)r = 6/d. However, with a large top riser,\n",
      "the cube loses its top surface for the purpose of heat dissipation.\"\"\"%Vr_\n",
      "Rc_ = 5/a\n",
      "dmin_=6/Rc_\n",
      "print \"d should be greater than or equal to %d cm\"%dmin_\n",
      "Vr__ = (pi/4)*dmin_**2 *floor(h)\n",
      "print \"The riser volume with minimum diameter is %d cm**3\"%Vr__"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(A/V)r=0.636422, (A/V)c=0.4\n",
        "Hence Rr is greater than Rc\n",
        "With minimum value of d Vr=2650 cm**3 .\n",
        "This valume is much more than the minimum Vr necessary.\n",
        "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2\n",
        "and again (A/V)r = 6/d. However, with a large top riser,\n",
        "the cube loses its top surface for the purpose of heat dissipation.\n",
        "d should be greater than or equal to 18 cm\n",
        "The riser volume with minimum diameter is 2290 cm**3\n"
       ]
      }
     ],
     "prompt_number": 8
    }
   ],
   "metadata": {}
  }
 ]
}