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diff --git a/TRANSPORT_PROCESSES_AND_UNIT_OPERATIONS/GeankoplisChapter14.ipynb b/TRANSPORT_PROCESSES_AND_UNIT_OPERATIONS/GeankoplisChapter14.ipynb
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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:9dc557e244e58a44c157bc89005c7dae2dc1b963927f29d131f60895891071ce"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 14: Mechanical-Physical Seperation Processes"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.2-1, Page number 810 "

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Evaluation of Filtration Constants for Constant-Pressure Filtration\n",

+      "#Variable declaration\n",

+      "A = 0.0439                 #Filter area of the plate and frame press (m2)\n",

+      "P = -338000.               #Pressure applied accross yhe plates (N/m2)\n",

+      "Cs = 23.47                 #Slurry concentration (kg/m3)\n",

+      "mu = 0.0008937              #Viscosity of water in SI units (kg/m.s)\n",

+      "\n",

+      "#Calculation\n",

+      "from numpy import arange,array,ones,linalg, divide\n",

+      "from pylab import plot,show\n",

+      "\n",

+      "xi = arange(0.0,0.00501,0.00005)\n",

+      "t = [4.4,9.5,16.3,24.6,34.7,46.1,59.0,73.6,89.4,107.3]\n",

+      "\n",

+      "# linearly generated sequence\n",

+      "y = [0.498e-3,1.0e-3,1.501e-3,2.0e-3,2.498e-3,3.002e-3,3.506e-3,4.004e-3,4.502e-3,5.009e-3]\n",

+      "yy = divide(t,y)\n",

+      "Aa = array([y, ones(10)])\n",

+      "w = linalg.lstsq(Aa.T,yy)[0] # obtaining the parameters\n",

+      "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",

+      "# plotting the line\n",

+      "line = w[0]*xi+w[1] # regression line\n",

+      "plot(xi,line,'r-',y,yy,'o')\n",

+      "ylabel('$t/V^2$')\n",

+      "xlabel('$V$')\n",

+      "xlim(0.0,5.1e-3)\n",

+      "ylim(0.0,22e3)\n",

+      "show()\n",

+      "Kp = 2.0*w[0]\n",

+      "B = w[1]\n",

+      "\n",

+      "alpha = (Kp*(A**2*(-P)))/(mu*Cs)\n",

+      "Rm = (B*(A*(-P)))/(mu)\n",

+      "#Result\n",

+      "print 'The value of constant alpha: %4.2e m/kg'%(alpha)\n",

+      "print 'The value of constant Rm: %4.3e 1/m'%(Rm)\n",

+      "print 'The values of alpha and Rm are more accurate than book the code utilises built in function'\n",

+      "print 'for regression, which gives better results than from graphical slope and intercept estimates'"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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EbKPQERER2/hs6AQHB9OvXz8GDBhAZGQkAGVlZcTExNCjRw9iY2M5cuSI5/UL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\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5fd7950>"

+       ]

+      },

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The value of constant alpha: 1.79e+11 m/kg\n",

+        "The value of constant Rm: 1.126e+11 1/m\n",

+        "The values of alpha and Rm are more accurate than book the code utilises built in function\n",

+        "for regression, which gives better results than from graphical slope and intercept estimates\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.2-2, Page number 811"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Time Required to perform a Filtration\n",

+      "from numpy import arange,array,ones,linalg, divide\n",

+      "from pylab import plot,show\n",

+      "\n",

+      "#Variable declaration\n",

+      "A = 0.873                  #Filter area of each frame (m2)\n",

+      "C = 0.0439                 #Proportionality constant\n",

+      "V = 3.37                   #Volume of filtrate (m3)\n",

+      "xi = arange(0.0,0.00501,0.00005)\n",

+      "t = [4.4,9.5,16.3,24.6,34.7,46.1,59.0,73.6,89.4,107.3]\n",

+      "y = [0.498e-3,1.0e-3,1.501e-3,2.0e-3,2.498e-3,3.002e-3,3.506e-3,4.004e-3,4.502e-3,5.009e-3]\n",

+      "\n",

+      "#Calculation\n",

+      "        # linearly generated sequence\n",

+      "yy = divide(t,y)\n",

+      "Aa = array([y, ones(10)])\n",

+      "w = linalg.lstsq(Aa.T,yy)[0] # obtaining the parameters\n",

+      "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",

+      "# plotting the line\n",

+      "line = w[0]*xi+w[1] # regression line\n",

+      "plot(xi,line,'r-',y,yy,'o')\n",

+      "ylabel('$t/V^2$')\n",

+      "xlabel('$V$')\n",

+      "xlim(0.0,5.1e-3)\n",

+      "ylim(0.0,22e3)\n",

+      "show()\n",

+      "Kp = 2.0*w[0]\n",

+      "B = w[1]\n",

+      "\n",

+      "At = A*20.                 #Total area of the 20 frames (m2)\n",

+      "Kpc = Kp*(C/At)**2         #Corrected slope (s/m6)\n",

+      "Bc = B*(C/At)              #Corrected intercept (s/m3)\n",

+      "#t = Kpc*V**2/2 + Bc*V\n",

+      "t = Kpc*V**2/2 + Bc*V\n",

+      "\n",

+      "#Result\n",

+      "print 'The time required to recover the filtrate is %5.2f s'%t\n",

+      "print 'The values slope and intercept of plot of t/v2 vs V is more accurate than book the code utilises '\n",

+      "print 'built in function for regression, which gives better results than from graphical estimates; Hence'\n",

+      "print 'time required to recover 3.37 m3 is different than given in book'"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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EbKPQERER2/hs6AQHB9OvXz8GDBhAZGQkAGVlZcTExNCjRw9iY2M5cuSI5/UL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\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5fd7eb0>"

+       ]

+      },

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The time required to recover the filtrate is 264.61 s\n",

+        "The values slope of plot of t/v2 vs V is more accurate than book the code utilises built in function\n",

+        "for regression, which gives better results than from graphical slope and intercept estimates; Hence\n",

+        "time required to recover 3.37 m3 is different than given in book\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.2-3, Page number 813"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Rate of Washing and Total Filter Cycle time\n",

+      "#Variable declaration\n",

+      "Kp = 37.93                 #Slope of the line multiplied by 2 (s/m6) \n",

+      "B = 16.10                  #Determined Intercept from data plotted (s/m3)\n",

+      "Vf = 3.37                  #Volume of the Flitrate (m3)\n",

+      "Vw = 0.337                 #Volume of wash water (m3)\n",

+      "Tc = 20.                   #Filter cleaning time (min)\n",

+      "Tr = 269.7                 #Filtrate recovry time (s)\n",

+      "#Calculation\n",

+      "#(dV/dt) = (1/4)*(1/(Kp*Vf + B))\n",

+      "R = (1./4.)*(1./((Kp*Vf) + B))\n",

+      "#Tw = (Volume of wash water)/(Rate of washing)\n",

+      "#Tw = Vw/(R)\n",

+      "Tw = Vw/(R)\n",

+      "#Total filtration cycle = Filtrate recovry time + Time of washing + Filter cleaning time\n",

+      "#Tt = Tr/60 + Tw/60 + Tc\n",

+      "Tt = Tr/60. + Tw/60. + 20.\n",

+      "#Result\n",

+      "print \"The time of washing is\",round(Tw,1),\"s\"\n",

+      "print \"The total filtration cycle is\",round(Tt,2),\"min\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The time of washing is 194.0 s\n",

+        "The total filtration cycle is 27.73 min\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.2-4, Page number 814 "

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Filtration in a continuous Rotary drumn\n",

+      "#Variable declaration\n",

+      "Rho = 996.9                 #Density of water (kg/m3)\n",

+      "Cx = 0.191                  #Solids concentration in slurry (kg)\n",

+      "m = 2.\n",

+      "P = -67.e3                  #Pressure drop in Filter (Pa)\n",

+      "M = 0.778                   #Mass of slurry (kg)\n",

+      "tc = 250.                   #Filter cycle time (s)\n",

+      "mu = 0.8937e-3              #Viscosity of slurry (Pa.s)\n",

+      "f = 0.33                    #Fraction of drum submerged\n",

+      "#Calculation\n",

+      "#Cs = (Rho*Cx)/(1-m*Cx)\n",

+      "Cs = (Rho*Cx)/(1-m*Cx)\n",

+      "alpha = (4.37e9)*(-P)**0.3\n",

+      "\n",

+      "R = M*Cx/Cs\n",

+      "\n",

+      "A = R*((tc*mu*alpha*Cs)/(2*f*(-P)))**0.5\n",

+      "#Result\n",

+      "\n",

+      "print \"The Filter Area needed to filter the slurry is \",round(A,2),\"m2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Filter Area needed to filter the slurry is  6.66 m2\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.3-1, Page number 818  "

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Filtration in a continuous Rotary drumn\n",

+      "#Variable declaration\n",

+      "Dp = 2.0e-5                     #Diameter of the oil droplet (m)\n",

+      "rhooil = 900.                   #Density of oil (kg/m3)\n",

+      "rhoair = 1.137                  #Density of air (kg/m3)\n",

+      "muair = 1.9e-5                  #Viscosity of air (Pa.s)\n",

+      "vt = 0.305                      #Assumed terminal settling velocity (m/s)\n",

+      "\n",

+      "#Calculation\n",

+      "eR = 1.\n",

+      "i = 0\n",

+      "while eR>=0.00001:\n",

+      "    i= i+1\n",

+      "    Nre = Dp*vt*rhoair/muair\n",

+      "    Cd = 24./Nre\n",

+      "    vtnew = sqrt(4.*(rhooil-rhoair)*9.81*Dp/(3*Cd*rhoair))\n",

+      "    eR = abs((vt-vtnew)/vt)\n",

+      "    vt = vtnew\n",

+      "#Result\n",

+      "print 'Reynolds number %4.3f'%Nre\n",

+      "print \"The calculated terminal settling velocity is\",round(vt,4),\"m/s\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Reynolds number 0.012\n",

+        "The calculated terminal settling velocity is 0.0103 m/s\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.3-2, Page number 820"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Hindered Settling of Spheres\n",

+      "\n",

+      "#Variable declaration\n",

+      "D = 1.554e-4                   #Diameter of the glass sphere (m)\n",

+      "rhoglass = 2467.               #Density of glass (kg/m3)\n",

+      "rhowater = 998.                #Density of water (kg/m3)\n",

+      "muwater = 1.005e-3             #Viscosity of water (Pa.s)\n",

+      "P_w = 40.                      #Percentage of water in mixture  \n",

+      "P_s = 60.                      #Percentage of solids in mixture \n",

+      "g = 9.807                      #Gravitational acceleration (m/s2)\n",

+      "#Calculation\n",

+      "e = (P_w/rhowater)/(P_w/rhowater + P_s/rhoglass)\n",

+      "rho_m = e*rhowater + (1 - e)*rhoglass\n",

+      "fi = 1/10**(1.82*(1-e))\n",

+      "vt = (g*D**2*(rhoglass - rhowater)*e**2*fi)/(18.*muwater)\n",

+      "N_re = (D*vt*rho_m*fi)/(muwater*e)\n",

+      "#Result\n",

+      "print 'The terminal velocity of the glass ball is %6.3e'%(vt),\"m/s\"\n",

+      "print \"The Reynolds number is\",round(N_re,3)\n",

+      "if N_re <= 1.:\n",

+      "    print 'Hence, the settling is in laminar range'"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The terminal velocity of the glass ball is 1.530e-03 m/s\n",

+        "The Reynolds number is 0.121\n",

+        "Hence, the settling is in laminar range\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.3-3, Page number 824"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Separation of a Mixture of Silica and Galena\n",

+      "\n",

+      "#Variable declaration\n",

+      "Dpl = 5.21e-6              #Size of the smallest particle (m)\n",

+      "Dpb = 2.50e-5              #Size of the largest particle  (m)       \n",

+      "rhoA = 7500.               #Density of Galena (kg/m3)\n",

+      "rhoB = 2650.               #Densiy of Silica (kg/m3)\n",

+      "rhol = 998.                #Density of Water (kg/m3)\n",

+      "mu = 1.005e-3              #Viscosity of Water (Pa.s)\n",

+      "g = 9.807                  #Gravitational accelertion (m/s2)\n",

+      "#Calculation\n",

+      "vtA =g*Dpb**2*(rhoA-rhol)/(18*mu)\n",

+      "Nre = Dpb*vtA*rhol/mu\n",

+      "DpA3 = Dpb*((rhoB-rhol)/(rhoA-rhol))**0.5\n",

+      "DpB2 = Dpl/((rhoB-rhol)/(rhoA-rhol))**0.5\n",

+      "\n",

+      "print 'The size range for first fraction of pure A Galena is as follows:\\n%5.3e to %5.3e' %(DpA3,Dpb)\n",

+      "print 'The size range for first fraction of pure B Silica is as follows:\\n%5.3e to %5.3e' %(Dpl,DpB2)\n",

+      "print 'The mixed fraction ranges is as follows:\\n%5.3e to %5.3e and \\n%5.3e to %5.3e' %(DpB2,Dpb,Dpl,DpA3) "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The size range for first fraction of pure A Galena is as follows:\n",

+        "1.260e-05 to 2.500e-05\n",

+        "The size range for first fraction of pure B Silica is as follows:\n",

+        "5.210e-06 to 1.034e-05\n",

+        "The mixed fraction ranges is as follows:\n",

+        "1.034e-05 to 2.500e-05 and \n",

+        "5.210e-06 to 1.260e-05\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.4-1, Page number 831"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Force in a centrifuge\n",

+      "\n",

+      "#Variable declaration\n",

+      "r1 = 0.1016              #Radius of the bowl in case A (m)\n",

+      "r2 = 0.2032              #Radius of the bowl in case B (m)\n",

+      "N = 1000.                #Revolution per minute  \n",

+      "g = 9.807                #Gravitational acceleration (m/s2)\n",

+      "\n",

+      "#Calculation\n",

+      "FcbyFg1 = r1*(2*pi*N/60)**2/g\n",

+      "FcbyFg2 = r2*(2*pi*N/60)**2/g\n",

+      "print \"Centrifugal force developed in case A:\", round(FcbyFg1,1),\"g's\"\n",

+      "print \"Centrifugal force developed in case B:\", round(FcbyFg2,1),\"g's\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Centrifugal force developed in case A: 113.6 g's\n",

+        "Centrifugal force developed in case B: 227.2 g's\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.4-2, Page number 833"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Settling in a centrifuge\n",

+      "\n",

+      "#Variable declaration\n",

+      "rhop = 1461.                  #Density of the viscous solution (kg/m3)\n",

+      "rhol = 801.                   #Density of the solution (kg/m3)\n",

+      "mu = 100e-3                   #Viscosity of the solution (Pa.s)\n",

+      "r1 = 0.00716                  #Inner radius of the bowl (m)\n",

+      "r2 = 0.02225                  #Outer radius of the bowl (m)\n",

+      "b = 0.1970                    #Height of the bowl (m)\n",

+      "N = 23000                     #Revolutions per minute\n",

+      "qc = 0.002832                 #Flowrate of the particles (m3/h)\n",

+      "\n",

+      "#Calculations\n",

+      "omegha = 2*pi*N/60\n",

+      "V = pi*b*(r2**2-r1**2)\n",

+      "qc = qc/3600\n",

+      "Dpc = sqrt(qc*18*mu*log(2*r2/(r1+r2))/(V*omegha**2*(rhop-rhol)))\n",

+      "vt = omegha**2*r2*Dpc**2*(rhop-rhol)/(18*mu)\n",

+      "Nre = Dpc*vt*rhol/mu\n",

+      "tT = V/qc\n",

+      "print \"The critical particle diameter of the largest particle in the exit stream is %5.3e m or %4.4f um\"%(Dpc,Dpc*1e6)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The critical particle diameter of the largest particle in the exit stream is 7.468e-07 m or 0.7468 um\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.4-3, Page number 836"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Location of interface in a centriguge\n",

+      "\n",

+      "#Variable declaration\n",

+      "rhooil = 919.5               #Density of oil phase (kg/m3)\n",

+      "rhoaqe = 980.1               #Density of aqueous phase (kg/m3)\n",

+      "r1 = 10.160                  #Radius for the overflow Light liquid (mm)\n",

+      "bh = 10.414                  #Radius for the overflow Heavy liquid (mm)\n",

+      "\n",

+      "#Calculations\n",

+      "r2 = sqrt((rhoaqe*bh**2-rhooil*r1**2)/(rhoaqe-rhooil))\n",

+      "\n",

+      "#Results\n",

+      "print \"Location of interface:\", round(r2,3),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Location of interface: 13.702 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 14.5-1 Page No. 842"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Power to Crust Iron Ore by Bond's Theory\n",

+      "\n",

+      "#Variable Declaration\n",

+      "T = 10.              #Feed Rate of Ore, ton/hr\n",

+      "Df = 3.              #Size of Feed , in\n",

+      "Dp = 1./8           #Size of product, in \n",

+      "\n",

+      "#Calculations\n",

+      "Ei = 12.68          #Work index for Iron Ore\n",

+      "T = T/60\n",

+      "Dp = Dp/12\n",

+      "Df = Df/12\n",

+      "\n",

+      "P= 1.46*T*Ei*(1./sqrt(Dp) - 1./sqrt(Df))\n",

+      "\n",

+      "#Result\n",

+      "print 'Power required %3.1f hp'%(P)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Power required 24.1 hp\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file